Solve 4x^2-8=10x

$$4 x ^ { 2 } - 8 = 10 x$$

Solve for x

$$x = \frac{\sqrt{57} + 5}{4} \approx 3.137458609$$

$$x=\frac{5-\sqrt{57}}{4}\approx -0.637458609$$

Steps Using the Quadratic Formula

Subtract 10x from both sides.

$$4x^{2}-8-10x=0$$

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.

$$4x^{2}-10x-8=0$$

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

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

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-10\right)±\sqrt{100-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-\left(-10\right)±\sqrt{100+128}}{2\times 4}$$

Add 100 to 128.

$$x=\frac{-\left(-10\right)±\sqrt{228}}{2\times 4}$$

Take the square root of 228.

$$x=\frac{-\left(-10\right)±2\sqrt{57}}{2\times 4}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{57}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{10±2\sqrt{57}}{8}$$

Now solve the equation x=\frac{10±2\sqrt{57}}{8} when ± is plus. Add 10 to 2\sqrt{57}.

$$x=\frac{2\sqrt{57}+10}{8}$$

Divide 10+2\sqrt{57} by 8.

$$x=\frac{\sqrt{57}+5}{4}$$

Now solve the equation x=\frac{10±2\sqrt{57}}{8} when ± is minus. Subtract 2\sqrt{57} from 10.

$$x=\frac{10-2\sqrt{57}}{8}$$

Divide 10-2\sqrt{57} by 8.

$$x=\frac{5-\sqrt{57}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{57}+5}{4} x=\frac{5-\sqrt{57}}{4}$$

Steps for Completing the Square

Subtract 10x from both sides.

$$4x^{2}-8-10x=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$4x^{2}-10x=8$$

Divide both sides by 4.

$$\frac{4x^{2}-10x}{4}=\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{2}x=\frac{8}{4}$$

Divide 8 by 4.

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

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.

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

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

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

Add 2 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{57}{16}$$

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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{57}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{57}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{57}}{4} x-\frac{5}{4}=-\frac{\sqrt{57}}{4}$$

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

$$x=\frac{\sqrt{57}+5}{4} x=\frac{5-\sqrt{57}}{4}$$