Solve 5x^2-20x+12=x^2+7x-6

$$5 x ^ { 2 } - 20 x + 12 = x ^ { 2 } + 7 x - 6$$

Solve for x

$$x=\frac{3}{4}=0.75$$

$$x=6$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$5x^{2}-20x+12-x^{2}=7x-6$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-20x+12=7x-6$$

Subtract 7x from both sides.

$$4x^{2}-20x+12-7x=-6$$

Combine -20x and -7x to get -27x.

$$4x^{2}-27x+12=-6$$

Add 6 to both sides.

$$4x^{2}-27x+12+6=0$$

Add 12 and 6 to get 18.

$$4x^{2}-27x+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+18. To find a and b, set up a system to be solved.

$$a+b=-27 ab=4\times 18=72$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 72.

$$-1,-72 -2,-36 -3,-24 -4,-18 -6,-12 -8,-9$$

Calculate the sum for each pair.

$$-1-72=-73 -2-36=-38 -3-24=-27 -4-18=-22 -6-12=-18 -8-9=-17$$

The solution is the pair that gives sum -27.

$$a=-24 b=-3$$

Rewrite 4x^{2}-27x+18 as \left(4x^{2}-24x\right)+\left(-3x+18\right).

$$\left(4x^{2}-24x\right)+\left(-3x+18\right)$$

Factor out 4x in the first and -3 in the second group.

$$4x\left(x-6\right)-3\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(4x-3\right)$$

To find equation solutions, solve x-6=0 and 4x-3=0.

$$x=6 x=\frac{3}{4}$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$5x^{2}-20x+12-x^{2}=7x-6$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-20x+12=7x-6$$

Subtract 7x from both sides.

$$4x^{2}-20x+12-7x=-6$$

Combine -20x and -7x to get -27x.

$$4x^{2}-27x+12=-6$$

Add 6 to both sides.

$$4x^{2}-27x+12+6=0$$

Add 12 and 6 to get 18.

$$4x^{2}-27x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -27 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 4\times 18}}{2\times 4}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 4\times 18}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-27\right)±\sqrt{729-16\times 18}}{2\times 4}$$

Multiply -16 times 18.

$$x=\frac{-\left(-27\right)±\sqrt{729-288}}{2\times 4}$$

Add 729 to -288.

$$x=\frac{-\left(-27\right)±\sqrt{441}}{2\times 4}$$

Take the square root of 441.

$$x=\frac{-\left(-27\right)±21}{2\times 4}$$

The opposite of -27 is 27.

$$x=\frac{27±21}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{27±21}{8}$$

Now solve the equation x=\frac{27±21}{8} when ± is plus. Add 27 to 21.

$$x=\frac{48}{8}$$

Divide 48 by 8.

$$x=6$$

Now solve the equation x=\frac{27±21}{8} when ± is minus. Subtract 21 from 27.

$$x=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{4}$$

The equation is now solved.

$$x=6 x=\frac{3}{4}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$5x^{2}-20x+12-x^{2}=7x-6$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-20x+12=7x-6$$

Subtract 7x from both sides.

$$4x^{2}-20x+12-7x=-6$$

Combine -20x and -7x to get -27x.

$$4x^{2}-27x+12=-6$$

Subtract 12 from both sides.

$$4x^{2}-27x=-6-12$$

Subtract 12 from -6 to get -18.

$$4x^{2}-27x=-18$$

Divide both sides by 4.

$$\frac{4x^{2}-27x}{4}=-\frac{18}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{27}{4}x=-\frac{18}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{27}{4}x=-\frac{9}{2}$$

Divide -\frac{27}{4}, the coefficient of the x term, by 2 to get -\frac{27}{8}. Then add the square of -\frac{27}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{27}{4}x+\left(-\frac{27}{8}\right)^{2}=-\frac{9}{2}+\left(-\frac{27}{8}\right)^{2}$$

Square -\frac{27}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{27}{4}x+\frac{729}{64}=-\frac{9}{2}+\frac{729}{64}$$

Add -\frac{9}{2} to \frac{729}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{27}{4}x+\frac{729}{64}=\frac{441}{64}$$

Factor x^{2}-\frac{27}{4}x+\frac{729}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{8}\right)^{2}=\frac{441}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{8}\right)^{2}}=\sqrt{\frac{441}{64}}$$

Simplify.

$$x-\frac{27}{8}=\frac{21}{8} x-\frac{27}{8}=-\frac{21}{8}$$

Add \frac{27}{8} to both sides of the equation.

$$x=6 x=\frac{3}{4}$$