Solve -x^2+14x-13=2x^2+6x-18-6x

$$- x ^ { 2 } + 14 x - 13 = 2 x ^ { 2 } + 6 x - 18 - 6 x$$

Solve for x

$$x=5$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using Factoring By Grouping

Combine 6x and -6x to get 0.

$$-x^{2}+14x-13=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$-x^{2}+14x-13-2x^{2}=-18$$

Add 18 to both sides.

$$-x^{2}+14x-13-2x^{2}+18=0$$

Add -13 and 18 to get 5.

$$-x^{2}+14x+5-2x^{2}=0$$

Combine -x^{2} and -2x^{2} to get -3x^{2}.

$$-3x^{2}+14x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=14 ab=-3\times 5=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 14.

$$a=15 b=-1$$

Rewrite -3x^{2}+14x+5 as \left(-3x^{2}+15x\right)+\left(-x+5\right).

$$\left(-3x^{2}+15x\right)+\left(-x+5\right)$$

Factor out 3x in -3x^{2}+15x.

$$3x\left(-x+5\right)-x+5$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(3x+1\right)$$

To find equation solutions, solve -x+5=0 and 3x+1=0.

$$x=5 x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Combine 6x and -6x to get 0.

$$-x^{2}+14x-13=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$-x^{2}+14x-13-2x^{2}=-18$$

Add 18 to both sides.

$$-x^{2}+14x-13-2x^{2}+18=0$$

Add -13 and 18 to get 5.

$$-x^{2}+14x+5-2x^{2}=0$$

Combine -x^{2} and -2x^{2} to get -3x^{2}.

$$-3x^{2}+14x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 14 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-14±\sqrt{196+12\times 5}}{2\left(-3\right)}$$

Multiply 12 times 5.

$$x=\frac{-14±\sqrt{196+60}}{2\left(-3\right)}$$

Add 196 to 60.

$$x=\frac{-14±\sqrt{256}}{2\left(-3\right)}$$

Take the square root of 256.

$$x=\frac{-14±16}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-14±16}{-6}$$

Now solve the equation x=\frac{-14±16}{-6} when ± is plus. Add -14 to 16.

$$x=\frac{2}{-6}$$

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

$$x=-\frac{1}{3}$$

Now solve the equation x=\frac{-14±16}{-6} when ± is minus. Subtract 16 from -14.

$$x=-\frac{30}{-6}$$

Divide -30 by -6.

$$x=5$$

The equation is now solved.

$$x=-\frac{1}{3} x=5$$

Steps for Completing the Square

Combine 6x and -6x to get 0.

$$-x^{2}+14x-13=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$-x^{2}+14x-13-2x^{2}=-18$$

Add 13 to both sides.

$$-x^{2}+14x-2x^{2}=-18+13$$

Add -18 and 13 to get -5.

$$-x^{2}+14x-2x^{2}=-5$$

Combine -x^{2} and -2x^{2} to get -3x^{2}.

$$-3x^{2}+14x=-5$$

Divide both sides by -3.

$$\frac{-3x^{2}+14x}{-3}=-\frac{5}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{14}{-3}x=-\frac{5}{-3}$$

Divide 14 by -3.

$$x^{2}-\frac{14}{3}x=-\frac{5}{-3}$$

Divide -5 by -3.

$$x^{2}-\frac{14}{3}x=\frac{5}{3}$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=\frac{5}{3}+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{5}{3}+\frac{49}{9}$$

Add \frac{5}{3} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{64}{9}$$

Factor x^{2}-\frac{14}{3}x+\frac{49}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{7}{3}=\frac{8}{3} x-\frac{7}{3}=-\frac{8}{3}$$

Add \frac{7}{3} to both sides of the equation.

$$x=5 x=-\frac{1}{3}$$