Solve 8=-5x^2+30x+35

$$8 = - 5 x ^ { 2 } + 30 x + 35$$

Solve for x

$$x=\frac{6\sqrt{10}}{5}+3\approx 6.794733192$$

$$x=-\frac{6\sqrt{10}}{5}+3\approx -0.794733192$$

Steps Using the Quadratic Formula

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

$$-5x^{2}+30x+35=8$$

Subtract 8 from both sides.

$$-5x^{2}+30x+35-8=0$$

Subtract 8 from 35 to get 27.

$$-5x^{2}+30x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 30 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-5\right)\times 27}}{2\left(-5\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-5\right)\times 27}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-30±\sqrt{900+20\times 27}}{2\left(-5\right)}$$

Multiply 20 times 27.

$$x=\frac{-30±\sqrt{900+540}}{2\left(-5\right)}$$

Add 900 to 540.

$$x=\frac{-30±\sqrt{1440}}{2\left(-5\right)}$$

Take the square root of 1440.

$$x=\frac{-30±12\sqrt{10}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-30±12\sqrt{10}}{-10}$$

Now solve the equation x=\frac{-30±12\sqrt{10}}{-10} when ± is plus. Add -30 to 12\sqrt{10}.

$$x=\frac{12\sqrt{10}-30}{-10}$$

Divide -30+12\sqrt{10} by -10.

$$x=-\frac{6\sqrt{10}}{5}+3$$

Now solve the equation x=\frac{-30±12\sqrt{10}}{-10} when ± is minus. Subtract 12\sqrt{10} from -30.

$$x=\frac{-12\sqrt{10}-30}{-10}$$

Divide -30-12\sqrt{10} by -10.

$$x=\frac{6\sqrt{10}}{5}+3$$

The equation is now solved.

$$x=-\frac{6\sqrt{10}}{5}+3 x=\frac{6\sqrt{10}}{5}+3$$

Steps for Completing the Square

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

$$-5x^{2}+30x+35=8$$

Subtract 35 from both sides.

$$-5x^{2}+30x=8-35$$

Subtract 35 from 8 to get -27.

$$-5x^{2}+30x=-27$$

Divide both sides by -5.

$$\frac{-5x^{2}+30x}{-5}=-\frac{27}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{30}{-5}x=-\frac{27}{-5}$$

Divide 30 by -5.

$$x^{2}-6x=-\frac{27}{-5}$$

Divide -27 by -5.

$$x^{2}-6x=\frac{27}{5}$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=\frac{27}{5}+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=\frac{27}{5}+9$$

Add \frac{27}{5} to 9.

$$x^{2}-6x+9=\frac{72}{5}$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=\frac{72}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{\frac{72}{5}}$$

Simplify.

$$x-3=\frac{6\sqrt{10}}{5} x-3=-\frac{6\sqrt{10}}{5}$$

Add 3 to both sides of the equation.

$$x=\frac{6\sqrt{10}}{5}+3 x=-\frac{6\sqrt{10}}{5}+3$$