Solve 4x^2-x*2=28+8x

$$4 { x }^{ 2 } -x \times 2 = 28+8x$$

Solve for x

$$x = \frac{\sqrt{137} + 5}{4} \approx 4.176174978$$

$$x=\frac{5-\sqrt{137}}{4}\approx -1.676174978$$

Steps Using the Quadratic Formula

Subtract 28 from both sides.

$$4x^{2}-x\times 2-28=8x$$

Subtract 8x from both sides.

$$4x^{2}-x\times 2-28-8x=0$$

Multiply -1 and 2 to get -2.

$$4x^{2}-2x-28-8x=0$$

Combine -2x and -8x to get -10x.

$$4x^{2}-10x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -10 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 4\left(-28\right)}}{2\times 4}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 4\left(-28\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-10\right)±\sqrt{100-16\left(-28\right)}}{2\times 4}$$

Multiply -16 times -28.

$$x=\frac{-\left(-10\right)±\sqrt{100+448}}{2\times 4}$$

Add 100 to 448.

$$x=\frac{-\left(-10\right)±\sqrt{548}}{2\times 4}$$

Take the square root of 548.

$$x=\frac{-\left(-10\right)±2\sqrt{137}}{2\times 4}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{137}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{10±2\sqrt{137}}{8}$$

Now solve the equation x=\frac{10±2\sqrt{137}}{8} when ± is plus. Add 10 to 2\sqrt{137}.

$$x=\frac{2\sqrt{137}+10}{8}$$

Divide 10+2\sqrt{137} by 8.

$$x=\frac{\sqrt{137}+5}{4}$$

Now solve the equation x=\frac{10±2\sqrt{137}}{8} when ± is minus. Subtract 2\sqrt{137} from 10.

$$x=\frac{10-2\sqrt{137}}{8}$$

Divide 10-2\sqrt{137} by 8.

$$x=\frac{5-\sqrt{137}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{137}+5}{4} x=\frac{5-\sqrt{137}}{4}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$4x^{2}-x\times 2-8x=28$$

Multiply -1 and 2 to get -2.

$$4x^{2}-2x-8x=28$$

Combine -2x and -8x to get -10x.

$$4x^{2}-10x=28$$

Divide both sides by 4.

$$\frac{4x^{2}-10x}{4}=\frac{28}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{10}{4}\right)x=\frac{28}{4}$$

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

$$x^{2}-\frac{5}{2}x=\frac{28}{4}$$

Divide 28 by 4.

$$x^{2}-\frac{5}{2}x=7$$

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.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=7+\left(-\frac{5}{4}\right)^{2}$$

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

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=7+\frac{25}{16}$$

Add 7 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{137}{16}$$

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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{137}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{137}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{137}}{4} x-\frac{5}{4}=-\frac{\sqrt{137}}{4}$$

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

$$x=\frac{\sqrt{137}+5}{4} x=\frac{5-\sqrt{137}}{4}$$