1. Resolver \( x^{2}+x-6 \geq 0.2 \) \( -2 x^{2}+3 x+2<0.5 \). Resolve

Solve the equation \( x^{2}+x-6 \geq 0.2 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}+x-6\geq 0.2\) - step1: Move the expression to the left side: \(x^{2}+x-6-0.2\geq 0\) - step2: Subtract the numbers: \(x^{2}+x-6.2\geq 0\) - step3: Rewrite the expression: \(x^{2}+x-6.2=0\) - step4: Convert the decimal into a fraction: \(x^{2}+x-\frac{31}{5}=0\) - step5: Add or subtract both sides: \(x^{2}+x=\frac{31}{5}\) - step6: Add the same value to both sides: \(x^{2}+x+\frac{1}{4}=\frac{31}{5}+\frac{1}{4}\) - step7: Simplify the expression: \(\left(x+\frac{1}{2}\right)^{2}=\frac{129}{20}\) - step8: Simplify the expression: \(x+\frac{1}{2}=\pm \sqrt{\frac{129}{20}}\) - step9: Simplify the expression: \(x+\frac{1}{2}=\pm \frac{\sqrt{645}}{10}\) - step10: Separate into possible cases: \(\begin{align}&x+\frac{1}{2}=\frac{\sqrt{645}}{10}\\&x+\frac{1}{2}=-\frac{\sqrt{645}}{10}\end{align}\) - step11: Solve the equation: \(\begin{align}&x=\frac{\sqrt{645}-5}{10}\\&x=-\frac{\sqrt{645}+5}{10}\end{align}\) - step12: Determine the test intervals: \(\begin{align}&x<-\frac{\sqrt{645}+5}{10}\\&-\frac{\sqrt{645}+5}{10}\frac{\sqrt{645}-5}{10}\end{align}\) - step13: Choose a value: \(\begin{align}&x_{1}=-4\\&x_{2}=-1\\&x_{3}=3\end{align}\) - step14: Test the chosen value: \(\begin{align}&x<-\frac{\sqrt{645}+5}{10}\textrm{ }\textrm{is the solution}\\&-\frac{\sqrt{645}+5}{10}\frac{\sqrt{645}-5}{10}\textrm{ }\textrm{is the solution}\end{align}\) - step15: Include the critical value: \(\begin{align}&x\leq -\frac{\sqrt{645}+5}{10}\textrm{ }\textrm{is the solution}\\&x\geq \frac{\sqrt{645}-5}{10}\textrm{ }\textrm{is the solution}\end{align}\) - step16: The final solution is \(x \in \left(-\infty,-\frac{\sqrt{645}+5}{10}\right]\cup \left[\frac{\sqrt{645}-5}{10},+\infty\right):\) \(x \in \left(-\infty,-\frac{\sqrt{645}+5}{10}\right]\cup \left[\frac{\sqrt{645}-5}{10},+\infty\right)\) Solve the equation \( -2 x^{2}+3 x+2<0.5 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(-2x^{2}+3x+2<0.5\) - step1: Move the expression to the left side: \(-2x^{2}+3x+2-0.5<0\) - step2: Subtract the numbers: \(-2x^{2}+3x+1.5<0\) - step3: Rewrite the expression: \(-2x^{2}+3x+1.5=0\) - step4: Convert the decimal into a fraction: \(-2x^{2}+3x+\frac{3}{2}=0\) - step5: Add or subtract both sides: \(-2x^{2}+3x=-\frac{3}{2}\) - step6: Divide both sides: \(\frac{-2x^{2}+3x}{-2}=\frac{-\frac{3}{2}}{-2}\) - step7: Evaluate: \(x^{2}-\frac{3}{2}x=\frac{3}{4}\) - step8: Add the same value to both sides: \(x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{3}{4}+\frac{9}{16}\) - step9: Simplify the expression: \(\left(x-\frac{3}{4}\right)^{2}=\frac{21}{16}\) - step10: Simplify the expression: \(x-\frac{3}{4}=\pm \sqrt{\frac{21}{16}}\) - step11: Simplify the expression: \(x-\frac{3}{4}=\pm \frac{\sqrt{21}}{4}\) - step12: Separate into possible cases: \(\begin{align}&x-\frac{3}{4}=\frac{\sqrt{21}}{4}\\&x-\frac{3}{4}=-\frac{\sqrt{21}}{4}\end{align}\) - step13: Solve the equation: \(\begin{align}&x=\frac{\sqrt{21}+3}{4}\\&x=\frac{-\sqrt{21}+3}{4}\end{align}\) - step14: Determine the test intervals: \(\begin{align}&x<\frac{-\sqrt{21}+3}{4}\\&\frac{-\sqrt{21}+3}{4}\frac{\sqrt{21}+3}{4}\end{align}\) - step15: Choose a value: \(\begin{align}&x_{1}=-1\\&x_{2}=1\\&x_{3}=3\end{align}\) - step16: Test the chosen value: \(\begin{align}&x<\frac{-\sqrt{21}+3}{4}\textrm{ }\textrm{is the solution}\\&\frac{-\sqrt{21}+3}{4}\frac{\sqrt{21}+3}{4}\textrm{ }\textrm{is the solution}\end{align}\) - step17: The final solution is \(x \in \left(-\infty,\frac{-\sqrt{21}+3}{4}\right)\cup \left(\frac{\sqrt{21}+3}{4},+\infty\right):\) \(x \in \left(-\infty,\frac{-\sqrt{21}+3}{4}\right)\cup \left(\frac{\sqrt{21}+3}{4},+\infty\right)\) La solución de la primera desigualdad \( x^{2}+x-6 \geq 0.2 \) es \( x \in (-\infty,-\frac{\sqrt{645}+5}{10}]\cup [\frac{\sqrt{645}-5}{10},+\infty) \). La solución de la segunda desigualdad \(-2 x^{2}+3 x+2<0.5\) es \( x \in (-\infty,\frac{-\sqrt{21}+3}{4})\cup (\frac{\sqrt{21}+3}{4},+\infty) \).