Solve 2x^2-8+7x=(-3x)^2-4^2

$$2 x ^ { 2 } - 8 + 7 x = ( - 3 x ) ^ { 2 } - 4 ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{273}}{14}+\frac{1}{2}\approx 1.680193689$$

$$x=-\frac{\sqrt{273}}{14}+\frac{1}{2}\approx -0.680193689$$

Steps Using the Quadratic Formula

Expand \left(-3x\right)^{2}.

$$2x^{2}-8+7x=\left(-3\right)^{2}x^{2}-4^{2}$$

Calculate -3 to the power of 2 and get 9.

$$2x^{2}-8+7x=9x^{2}-4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$2x^{2}-8+7x=9x^{2}-16$$

Subtract 9x^{2} from both sides.

$$2x^{2}-8+7x-9x^{2}=-16$$

Combine 2x^{2} and -9x^{2} to get -7x^{2}.

$$-7x^{2}-8+7x=-16$$

Add 16 to both sides.

$$-7x^{2}-8+7x+16=0$$

Add -8 and 16 to get 8.

$$-7x^{2}+8+7x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-7x^{2}+7x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 7 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-7\right)\times 8}}{2\left(-7\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-7\right)\times 8}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-7±\sqrt{49+28\times 8}}{2\left(-7\right)}$$

Multiply 28 times 8.

$$x=\frac{-7±\sqrt{49+224}}{2\left(-7\right)}$$

Add 49 to 224.

$$x=\frac{-7±\sqrt{273}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-7±\sqrt{273}}{-14}$$

Now solve the equation x=\frac{-7±\sqrt{273}}{-14} when ± is plus. Add -7 to \sqrt{273}.

$$x=\frac{\sqrt{273}-7}{-14}$$

Divide -7+\sqrt{273} by -14.

$$x=-\frac{\sqrt{273}}{14}+\frac{1}{2}$$

Now solve the equation x=\frac{-7±\sqrt{273}}{-14} when ± is minus. Subtract \sqrt{273} from -7.

$$x=\frac{-\sqrt{273}-7}{-14}$$

Divide -7-\sqrt{273} by -14.

$$x=\frac{\sqrt{273}}{14}+\frac{1}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{273}}{14}+\frac{1}{2} x=\frac{\sqrt{273}}{14}+\frac{1}{2}$$

Steps for Completing the Square

Expand \left(-3x\right)^{2}.

$$2x^{2}-8+7x=\left(-3\right)^{2}x^{2}-4^{2}$$

Calculate -3 to the power of 2 and get 9.

$$2x^{2}-8+7x=9x^{2}-4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$2x^{2}-8+7x=9x^{2}-16$$

Subtract 9x^{2} from both sides.

$$2x^{2}-8+7x-9x^{2}=-16$$

Combine 2x^{2} and -9x^{2} to get -7x^{2}.

$$-7x^{2}-8+7x=-16$$

Add 8 to both sides.

$$-7x^{2}+7x=-16+8$$

Add -16 and 8 to get -8.

$$-7x^{2}+7x=-8$$

Divide both sides by -7.

$$\frac{-7x^{2}+7x}{-7}=-\frac{8}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\frac{7}{-7}x=-\frac{8}{-7}$$

Divide 7 by -7.

$$x^{2}-x=-\frac{8}{-7}$$

Divide -8 by -7.

$$x^{2}-x=\frac{8}{7}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{8}{7}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{8}{7}+\frac{1}{4}$$

Add \frac{8}{7} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{39}{28}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{39}{28}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{39}{28}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{273}}{14} x-\frac{1}{2}=-\frac{\sqrt{273}}{14}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{273}}{14}+\frac{1}{2} x=-\frac{\sqrt{273}}{14}+\frac{1}{2}$$