Solve 8x^2-5x+87=7x^2+5x

$$8 x ^ { 2 } - 5 x + 87 = 7 x ^ { 2 } + 5 x$$

Solve for x (complex solution)

$$x=5+\sqrt{62}i\approx 5+7.874007874i$$

$$x=-\sqrt{62}i+5\approx 5-7.874007874i$$

Steps Using the Quadratic Formula

Subtract 7x^{2} from both sides.

$$8x^{2}-5x+87-7x^{2}=5x$$

Combine 8x^{2} and -7x^{2} to get x^{2}.

$$x^{2}-5x+87=5x$$

Subtract 5x from both sides.

$$x^{2}-5x+87-5x=0$$

Combine -5x and -5x to get -10x.

$$x^{2}-10x+87=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 87 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 87}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 87}}{2}$$

Multiply -4 times 87.

$$x=\frac{-\left(-10\right)±\sqrt{100-348}}{2}$$

Add 100 to -348.

$$x=\frac{-\left(-10\right)±\sqrt{-248}}{2}$$

Take the square root of -248.

$$x=\frac{-\left(-10\right)±2\sqrt{62}i}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{62}i}{2}$$

Now solve the equation x=\frac{10±2\sqrt{62}i}{2} when ± is plus. Add 10 to 2i\sqrt{62}.

$$x=\frac{10+2\sqrt{62}i}{2}$$

Divide 10+2i\sqrt{62} by 2.

$$x=5+\sqrt{62}i$$

Now solve the equation x=\frac{10±2\sqrt{62}i}{2} when ± is minus. Subtract 2i\sqrt{62} from 10.

$$x=\frac{-2\sqrt{62}i+10}{2}$$

Divide 10-2i\sqrt{62} by 2.

$$x=-\sqrt{62}i+5$$

The equation is now solved.

$$x=5+\sqrt{62}i x=-\sqrt{62}i+5$$

Steps for Completing the Square

Subtract 7x^{2} from both sides.

$$8x^{2}-5x+87-7x^{2}=5x$$

Combine 8x^{2} and -7x^{2} to get x^{2}.

$$x^{2}-5x+87=5x$$

Subtract 5x from both sides.

$$x^{2}-5x+87-5x=0$$

Combine -5x and -5x to get -10x.

$$x^{2}-10x+87=0$$

Subtract 87 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-10x=-87$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-87+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-87+25$$

Add -87 to 25.

$$x^{2}-10x+25=-62$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=-62$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{-62}$$

Simplify.

$$x-5=\sqrt{62}i x-5=-\sqrt{62}i$$

Add 5 to both sides of the equation.

$$x=5+\sqrt{62}i x=-\sqrt{62}i+5$$