Solve 7-6(x-1)=7x-(2-6x)x=

$$7 - 6 ( x - 1 ) = 7 x - ( 2 - 6 x ) x =$$

Solve for x

$$x=\frac{\sqrt{433}-11}{12}\approx 0.817387671$$

$$x=\frac{-\sqrt{433}-11}{12}\approx -2.650721004$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -6 by x-1.

$$7-6x+6=7x-\left(2-6x\right)x$$

Add 7 and 6 to get 13.

$$13-6x=7x-\left(2-6x\right)x$$

Use the distributive property to multiply 2-6x by x.

$$13-6x=7x-\left(2x-6x^{2}\right)$$

To find the opposite of 2x-6x^{2}, find the opposite of each term.

$$13-6x=7x-2x-\left(-6x^{2}\right)$$

The opposite of -6x^{2} is 6x^{2}.

$$13-6x=7x-2x+6x^{2}$$

Combine 7x and -2x to get 5x.

$$13-6x=5x+6x^{2}$$

Subtract 5x from both sides.

$$13-6x-5x=6x^{2}$$

Combine -6x and -5x to get -11x.

$$13-11x=6x^{2}$$

Subtract 6x^{2} from both sides.

$$13-11x-6x^{2}=0$$

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.

$$-6x^{2}-11x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -11 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-6\right)\times 13}}{2\left(-6\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-6\right)\times 13}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-11\right)±\sqrt{121+24\times 13}}{2\left(-6\right)}$$

Multiply 24 times 13.

$$x=\frac{-\left(-11\right)±\sqrt{121+312}}{2\left(-6\right)}$$

Add 121 to 312.

$$x=\frac{-\left(-11\right)±\sqrt{433}}{2\left(-6\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{433}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{11±\sqrt{433}}{-12}$$

Now solve the equation x=\frac{11±\sqrt{433}}{-12} when ± is plus. Add 11 to \sqrt{433}.

$$x=\frac{\sqrt{433}+11}{-12}$$

Divide 11+\sqrt{433} by -12.

$$x=\frac{-\sqrt{433}-11}{12}$$

Now solve the equation x=\frac{11±\sqrt{433}}{-12} when ± is minus. Subtract \sqrt{433} from 11.

$$x=\frac{11-\sqrt{433}}{-12}$$

Divide 11-\sqrt{433} by -12.

$$x=\frac{\sqrt{433}-11}{12}$$

The equation is now solved.

$$x=\frac{-\sqrt{433}-11}{12} x=\frac{\sqrt{433}-11}{12}$$

Steps for Completing the Square

Use the distributive property to multiply -6 by x-1.

$$7-6x+6=7x-\left(2-6x\right)x$$

Add 7 and 6 to get 13.

$$13-6x=7x-\left(2-6x\right)x$$

Use the distributive property to multiply 2-6x by x.

$$13-6x=7x-\left(2x-6x^{2}\right)$$

To find the opposite of 2x-6x^{2}, find the opposite of each term.

$$13-6x=7x-2x-\left(-6x^{2}\right)$$

The opposite of -6x^{2} is 6x^{2}.

$$13-6x=7x-2x+6x^{2}$$

Combine 7x and -2x to get 5x.

$$13-6x=5x+6x^{2}$$

Subtract 5x from both sides.

$$13-6x-5x=6x^{2}$$

Combine -6x and -5x to get -11x.

$$13-11x=6x^{2}$$

Subtract 6x^{2} from both sides.

$$13-11x-6x^{2}=0$$

Subtract 13 from both sides. Anything subtracted from zero gives its negation.

$$-11x-6x^{2}=-13$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-6x^{2}-11x=-13$$

Divide both sides by -6.

$$\frac{-6x^{2}-11x}{-6}=-\frac{13}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\left(-\frac{11}{-6}\right)x=-\frac{13}{-6}$$

Divide -11 by -6.

$$x^{2}+\frac{11}{6}x=-\frac{13}{-6}$$

Divide -13 by -6.

$$x^{2}+\frac{11}{6}x=\frac{13}{6}$$

Divide \frac{11}{6}, the coefficient of the x term, by 2 to get \frac{11}{12}. Then add the square of \frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{6}x+\left(\frac{11}{12}\right)^{2}=\frac{13}{6}+\left(\frac{11}{12}\right)^{2}$$

Square \frac{11}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{6}x+\frac{121}{144}=\frac{13}{6}+\frac{121}{144}$$

Add \frac{13}{6} to \frac{121}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{6}x+\frac{121}{144}=\frac{433}{144}$$

Factor x^{2}+\frac{11}{6}x+\frac{121}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{12}\right)^{2}=\frac{433}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{12}\right)^{2}}=\sqrt{\frac{433}{144}}$$

Simplify.

$$x+\frac{11}{12}=\frac{\sqrt{433}}{12} x+\frac{11}{12}=-\frac{\sqrt{433}}{12}$$

Subtract \frac{11}{12} from both sides of the equation.

$$x=\frac{\sqrt{433}-11}{12} x=\frac{-\sqrt{433}-11}{12}$$