Solve 0x(2x-1)=8+x(2x+35)

$$0 x ( 2 x - 1 ) = 8 + x ( 2 x + 35 )$$

Solve for x

$$x=\frac{3\sqrt{129}-35}{4}\approx -0.231637481$$

$$x=\frac{-3\sqrt{129}-35}{4}\approx -17.268362519$$

Steps Using the Quadratic Formula

Anything times zero gives zero.

$$0=8+x\left(2x+35\right)$$

Use the distributive property to multiply x by 2x+35.

$$0=8+2x^{2}+35x$$

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

$$8+2x^{2}+35x=0$$

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.

$$2x^{2}+35x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 35 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-35±\sqrt{35^{2}-4\times 2\times 8}}{2\times 2}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\times 2\times 8}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-35±\sqrt{1225-8\times 8}}{2\times 2}$$

Multiply -8 times 8.

$$x=\frac{-35±\sqrt{1225-64}}{2\times 2}$$

Add 1225 to -64.

$$x=\frac{-35±\sqrt{1161}}{2\times 2}$$

Take the square root of 1161.

$$x=\frac{-35±3\sqrt{129}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-35±3\sqrt{129}}{4}$$

Now solve the equation x=\frac{-35±3\sqrt{129}}{4} when ± is plus. Add -35 to 3\sqrt{129}.

$$x=\frac{3\sqrt{129}-35}{4}$$

Now solve the equation x=\frac{-35±3\sqrt{129}}{4} when ± is minus. Subtract 3\sqrt{129} from -35.

$$x=\frac{-3\sqrt{129}-35}{4}$$

The equation is now solved.

$$x=\frac{3\sqrt{129}-35}{4} x=\frac{-3\sqrt{129}-35}{4}$$

Steps for Completing the Square

Anything times zero gives zero.

$$0=8+x\left(2x+35\right)$$

Use the distributive property to multiply x by 2x+35.

$$0=8+2x^{2}+35x$$

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

$$8+2x^{2}+35x=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

$$\frac{2x^{2}+35x}{2}=-\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{35}{2}x=-\frac{8}{2}$$

Divide -8 by 2.

$$x^{2}+\frac{35}{2}x=-4$$

Divide \frac{35}{2}, the coefficient of the x term, by 2 to get \frac{35}{4}. Then add the square of \frac{35}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{35}{2}x+\left(\frac{35}{4}\right)^{2}=-4+\left(\frac{35}{4}\right)^{2}$$

Square \frac{35}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{35}{2}x+\frac{1225}{16}=-4+\frac{1225}{16}$$

Add -4 to \frac{1225}{16}.

$$x^{2}+\frac{35}{2}x+\frac{1225}{16}=\frac{1161}{16}$$

Factor x^{2}+\frac{35}{2}x+\frac{1225}{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}.

$$\left(x+\frac{35}{4}\right)^{2}=\frac{1161}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{35}{4}\right)^{2}}=\sqrt{\frac{1161}{16}}$$

Simplify.

$$x+\frac{35}{4}=\frac{3\sqrt{129}}{4} x+\frac{35}{4}=-\frac{3\sqrt{129}}{4}$$

Subtract \frac{35}{4} from both sides of the equation.

$$x=\frac{3\sqrt{129}-35}{4} x=\frac{-3\sqrt{129}-35}{4}$$