Solve x^2-7x+12=20x^2+x+1

$${ x }^{ 2 } -7x+12=20 { x }^{ 2 } +x+1$$

Solve for x

$$x=-1$$

$$x=\frac{11}{19}\approx 0.578947368$$

Steps Using Factoring By Grouping

Subtract 20x^{2} from both sides.

$$x^{2}-7x+12-20x^{2}=x+1$$

Combine x^{2} and -20x^{2} to get -19x^{2}.

$$-19x^{2}-7x+12=x+1$$

Subtract x from both sides.

$$-19x^{2}-7x+12-x=1$$

Combine -7x and -x to get -8x.

$$-19x^{2}-8x+12=1$$

Subtract 1 from both sides.

$$-19x^{2}-8x+12-1=0$$

Subtract 1 from 12 to get 11.

$$-19x^{2}-8x+11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -19x^{2}+ax+bx+11. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-19\times 11=-209$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -209.

$$1,-209 11,-19$$

Calculate the sum for each pair.

$$1-209=-208 11-19=-8$$

The solution is the pair that gives sum -8.

$$a=11 b=-19$$

Rewrite -19x^{2}-8x+11 as \left(-19x^{2}+11x\right)+\left(-19x+11\right).

$$\left(-19x^{2}+11x\right)+\left(-19x+11\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(19x-11\right)-\left(19x-11\right)$$

Factor out common term 19x-11 by using distributive property.

$$\left(19x-11\right)\left(-x-1\right)$$

To find equation solutions, solve 19x-11=0 and -x-1=0.

$$x=\frac{11}{19} x=-1$$

Steps Using the Quadratic Formula

Subtract 20x^{2} from both sides.

$$x^{2}-7x+12-20x^{2}=x+1$$

Combine x^{2} and -20x^{2} to get -19x^{2}.

$$-19x^{2}-7x+12=x+1$$

Subtract x from both sides.

$$-19x^{2}-7x+12-x=1$$

Combine -7x and -x to get -8x.

$$-19x^{2}-8x+12=1$$

Subtract 1 from both sides.

$$-19x^{2}-8x+12-1=0$$

Subtract 1 from 12 to get 11.

$$-19x^{2}-8x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -19 for a, -8 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-19\right)\times 11}}{2\left(-19\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-19\right)\times 11}}{2\left(-19\right)}$$

Multiply -4 times -19.

$$x=\frac{-\left(-8\right)±\sqrt{64+76\times 11}}{2\left(-19\right)}$$

Multiply 76 times 11.

$$x=\frac{-\left(-8\right)±\sqrt{64+836}}{2\left(-19\right)}$$

Add 64 to 836.

$$x=\frac{-\left(-8\right)±\sqrt{900}}{2\left(-19\right)}$$

Take the square root of 900.

$$x=\frac{-\left(-8\right)±30}{2\left(-19\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±30}{2\left(-19\right)}$$

Multiply 2 times -19.

$$x=\frac{8±30}{-38}$$

Now solve the equation x=\frac{8±30}{-38} when ± is plus. Add 8 to 30.

$$x=\frac{38}{-38}$$

Divide 38 by -38.

$$x=-1$$

Now solve the equation x=\frac{8±30}{-38} when ± is minus. Subtract 30 from 8.

$$x=-\frac{22}{-38}$$

Reduce the fraction \frac{-22}{-38} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{19}$$

The equation is now solved.

$$x=-1 x=\frac{11}{19}$$

Steps for Completing the Square

Subtract 20x^{2} from both sides.

$$x^{2}-7x+12-20x^{2}=x+1$$

Combine x^{2} and -20x^{2} to get -19x^{2}.

$$-19x^{2}-7x+12=x+1$$

Subtract x from both sides.

$$-19x^{2}-7x+12-x=1$$

Combine -7x and -x to get -8x.

$$-19x^{2}-8x+12=1$$

Subtract 12 from both sides.

$$-19x^{2}-8x=1-12$$

Subtract 12 from 1 to get -11.

$$-19x^{2}-8x=-11$$

Divide both sides by -19.

$$\frac{-19x^{2}-8x}{-19}=-\frac{11}{-19}$$

Dividing by -19 undoes the multiplication by -19.

$$x^{2}+\left(-\frac{8}{-19}\right)x=-\frac{11}{-19}$$

Divide -8 by -19.

$$x^{2}+\frac{8}{19}x=-\frac{11}{-19}$$

Divide -11 by -19.

$$x^{2}+\frac{8}{19}x=\frac{11}{19}$$

Divide \frac{8}{19}, the coefficient of the x term, by 2 to get \frac{4}{19}. Then add the square of \frac{4}{19} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{19}x+\left(\frac{4}{19}\right)^{2}=\frac{11}{19}+\left(\frac{4}{19}\right)^{2}$$

Square \frac{4}{19} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{19}x+\frac{16}{361}=\frac{11}{19}+\frac{16}{361}$$

Add \frac{11}{19} to \frac{16}{361} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{19}x+\frac{16}{361}=\frac{225}{361}$$

Factor x^{2}+\frac{8}{19}x+\frac{16}{361}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{19}\right)^{2}=\frac{225}{361}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{19}\right)^{2}}=\sqrt{\frac{225}{361}}$$

Simplify.

$$x+\frac{4}{19}=\frac{15}{19} x+\frac{4}{19}=-\frac{15}{19}$$

Subtract \frac{4}{19} from both sides of the equation.

$$x=\frac{11}{19} x=-1$$