Solve 7x^2+2=30x-10

$$7 { x }^{ 2 } +2=30x-10$$

Solve for x

$$x = \frac{\sqrt{141} + 15}{7} \approx 3.839191727$$

$$x=\frac{15-\sqrt{141}}{7}\approx 0.446522559$$

Steps Using the Quadratic Formula

Subtract 30x from both sides.

$$7x^{2}+2-30x=-10$$

Add 10 to both sides.

$$7x^{2}+2-30x+10=0$$

Add 2 and 10 to get 12.

$$7x^{2}+12-30x=0$$

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.

$$7x^{2}-30x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -30 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 7\times 12}}{2\times 7}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 7\times 12}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-30\right)±\sqrt{900-28\times 12}}{2\times 7}$$

Multiply -28 times 12.

$$x=\frac{-\left(-30\right)±\sqrt{900-336}}{2\times 7}$$

Add 900 to -336.

$$x=\frac{-\left(-30\right)±\sqrt{564}}{2\times 7}$$

Take the square root of 564.

$$x=\frac{-\left(-30\right)±2\sqrt{141}}{2\times 7}$$

The opposite of -30 is 30.

$$x=\frac{30±2\sqrt{141}}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{30±2\sqrt{141}}{14}$$

Now solve the equation x=\frac{30±2\sqrt{141}}{14} when ± is plus. Add 30 to 2\sqrt{141}.

$$x=\frac{2\sqrt{141}+30}{14}$$

Divide 30+2\sqrt{141} by 14.

$$x=\frac{\sqrt{141}+15}{7}$$

Now solve the equation x=\frac{30±2\sqrt{141}}{14} when ± is minus. Subtract 2\sqrt{141} from 30.

$$x=\frac{30-2\sqrt{141}}{14}$$

Divide 30-2\sqrt{141} by 14.

$$x=\frac{15-\sqrt{141}}{7}$$

The equation is now solved.

$$x=\frac{\sqrt{141}+15}{7} x=\frac{15-\sqrt{141}}{7}$$

Steps for Completing the Square

Subtract 30x from both sides.

$$7x^{2}+2-30x=-10$$

Subtract 2 from both sides.

$$7x^{2}-30x=-10-2$$

Subtract 2 from -10 to get -12.

$$7x^{2}-30x=-12$$

Divide both sides by 7.

$$\frac{7x^{2}-30x}{7}=-\frac{12}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{30}{7}x=-\frac{12}{7}$$

Divide -\frac{30}{7}, the coefficient of the x term, by 2 to get -\frac{15}{7}. Then add the square of -\frac{15}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{30}{7}x+\left(-\frac{15}{7}\right)^{2}=-\frac{12}{7}+\left(-\frac{15}{7}\right)^{2}$$

Square -\frac{15}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{30}{7}x+\frac{225}{49}=-\frac{12}{7}+\frac{225}{49}$$

Add -\frac{12}{7} to \frac{225}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{30}{7}x+\frac{225}{49}=\frac{141}{49}$$

Factor x^{2}-\frac{30}{7}x+\frac{225}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{7}\right)^{2}=\frac{141}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{7}\right)^{2}}=\sqrt{\frac{141}{49}}$$

Simplify.

$$x-\frac{15}{7}=\frac{\sqrt{141}}{7} x-\frac{15}{7}=-\frac{\sqrt{141}}{7}$$

Add \frac{15}{7} to both sides of the equation.

$$x=\frac{\sqrt{141}+15}{7} x=\frac{15-\sqrt{141}}{7}$$