Solve 14=-10x^2+25x+6

$$14 = - 10 x ^ { 2 } + 25 x + 6$$

Solve for x

$$x=\frac{\sqrt{305}}{20}+\frac{5}{4}\approx 2.12321246$$

$$x=-\frac{\sqrt{305}}{20}+\frac{5}{4}\approx 0.37678754$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-10x^{2}+25x+6=14$$

Subtract 14 from both sides.

$$-10x^{2}+25x+6-14=0$$

Subtract 14 from 6 to get -8.

$$-10x^{2}+25x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 25 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-10\right)\left(-8\right)}}{2\left(-10\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-10\right)\left(-8\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-25±\sqrt{625+40\left(-8\right)}}{2\left(-10\right)}$$

Multiply 40 times -8.

$$x=\frac{-25±\sqrt{625-320}}{2\left(-10\right)}$$

Add 625 to -320.

$$x=\frac{-25±\sqrt{305}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-25±\sqrt{305}}{-20}$$

Now solve the equation x=\frac{-25±\sqrt{305}}{-20} when ± is plus. Add -25 to \sqrt{305}.

$$x=\frac{\sqrt{305}-25}{-20}$$

Divide -25+\sqrt{305} by -20.

$$x=-\frac{\sqrt{305}}{20}+\frac{5}{4}$$

Now solve the equation x=\frac{-25±\sqrt{305}}{-20} when ± is minus. Subtract \sqrt{305} from -25.

$$x=\frac{-\sqrt{305}-25}{-20}$$

Divide -25-\sqrt{305} by -20.

$$x=\frac{\sqrt{305}}{20}+\frac{5}{4}$$

The equation is now solved.

$$x=-\frac{\sqrt{305}}{20}+\frac{5}{4} x=\frac{\sqrt{305}}{20}+\frac{5}{4}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-10x^{2}+25x+6=14$$

Subtract 6 from both sides.

$$-10x^{2}+25x=14-6$$

Subtract 6 from 14 to get 8.

$$-10x^{2}+25x=8$$

Divide both sides by -10.

$$\frac{-10x^{2}+25x}{-10}=\frac{8}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{25}{-10}x=\frac{8}{-10}$$

Reduce the fraction \frac{25}{-10} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{5}{2}x=\frac{8}{-10}$$

Reduce the fraction \frac{8}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{2}x=-\frac{4}{5}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{4}{5}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{4}{5}+\frac{25}{16}$$

Add -\frac{4}{5} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{61}{80}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{61}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{61}{80}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{305}}{20} x-\frac{5}{4}=-\frac{\sqrt{305}}{20}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{\sqrt{305}}{20}+\frac{5}{4} x=-\frac{\sqrt{305}}{20}+\frac{5}{4}$$