Solve 4x^2-30x=12+10x

$$4 x ^ { 2 } - 30 x = 12 + 10 x$$

Solve for x

$$x=2\sqrt{7}+5\approx 10.291502622$$

$$x=5-2\sqrt{7}\approx -0.291502622$$

Steps Using the Quadratic Formula

Subtract 12 from both sides.

$$4x^{2}-30x-12=10x$$

Subtract 10x from both sides.

$$4x^{2}-30x-12-10x=0$$

Combine -30x and -10x to get -40x.

$$4x^{2}-40x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -40 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 4\left(-12\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-40\right)±\sqrt{1600-16\left(-12\right)}}{2\times 4}$$

Multiply -16 times -12.

$$x=\frac{-\left(-40\right)±\sqrt{1600+192}}{2\times 4}$$

Add 1600 to 192.

$$x=\frac{-\left(-40\right)±\sqrt{1792}}{2\times 4}$$

Take the square root of 1792.

$$x=\frac{-\left(-40\right)±16\sqrt{7}}{2\times 4}$$

The opposite of -40 is 40.

$$x=\frac{40±16\sqrt{7}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{40±16\sqrt{7}}{8}$$

Now solve the equation x=\frac{40±16\sqrt{7}}{8} when ± is plus. Add 40 to 16\sqrt{7}.

$$x=\frac{16\sqrt{7}+40}{8}$$

Divide 40+16\sqrt{7} by 8.

$$x=2\sqrt{7}+5$$

Now solve the equation x=\frac{40±16\sqrt{7}}{8} when ± is minus. Subtract 16\sqrt{7} from 40.

$$x=\frac{40-16\sqrt{7}}{8}$$

Divide 40-16\sqrt{7} by 8.

$$x=5-2\sqrt{7}$$

The equation is now solved.

$$x=2\sqrt{7}+5 x=5-2\sqrt{7}$$

Steps for Completing the Square

Subtract 10x from both sides.

$$4x^{2}-30x-10x=12$$

Combine -30x and -10x to get -40x.

$$4x^{2}-40x=12$$

Divide both sides by 4.

$$\frac{4x^{2}-40x}{4}=\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{40}{4}\right)x=\frac{12}{4}$$

Divide -40 by 4.

$$x^{2}-10x=\frac{12}{4}$$

Divide 12 by 4.

$$x^{2}-10x=3$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=3+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=3+25$$

Add 3 to 25.

$$x^{2}-10x+25=28$$

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}.

$$\left(x-5\right)^{2}=28$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{28}$$

Simplify.

$$x-5=2\sqrt{7} x-5=-2\sqrt{7}$$

Add 5 to both sides of the equation.

$$x=2\sqrt{7}+5 x=5-2\sqrt{7}$$