Solve 7left(x-3right)-5left(x^2-1right)=x^2-5left(x+2right)

$$7 \left( x-3 \right) -5 \left( { x }^{ 2 } -1 \right) = { x }^{ 2 } -5 \left( x+2 \right)$$

Solve for x

$$x=1$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 7 by x-3.

$$7x-21-5\left(x^{2}-1\right)=x^{2}-5\left(x+2\right)$$

Use the distributive property to multiply -5 by x^{2}-1.

$$7x-21-5x^{2}+5=x^{2}-5\left(x+2\right)$$

Add -21 and 5 to get -16.

$$7x-16-5x^{2}=x^{2}-5\left(x+2\right)$$

Use the distributive property to multiply -5 by x+2.

$$7x-16-5x^{2}=x^{2}-5x-10$$

Subtract x^{2} from both sides.

$$7x-16-5x^{2}-x^{2}=-5x-10$$

Combine -5x^{2} and -x^{2} to get -6x^{2}.

$$7x-16-6x^{2}=-5x-10$$

Add 5x to both sides.

$$7x-16-6x^{2}+5x=-10$$

Combine 7x and 5x to get 12x.

$$12x-16-6x^{2}=-10$$

Add 10 to both sides.

$$12x-16-6x^{2}+10=0$$

Add -16 and 10 to get -6.

$$12x-6-6x^{2}=0$$

Divide both sides by 6.

$$2x-1-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite -x^{2}+2x-1 as \left(-x^{2}+x\right)+\left(x-1\right).

$$\left(-x^{2}+x\right)+\left(x-1\right)$$

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

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

Factor out common term x-1 by using distributive property.

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

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

$$x=1 x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 7 by x-3.

$$7x-21-5\left(x^{2}-1\right)=x^{2}-5\left(x+2\right)$$

Use the distributive property to multiply -5 by x^{2}-1.

$$7x-21-5x^{2}+5=x^{2}-5\left(x+2\right)$$

Add -21 and 5 to get -16.

$$7x-16-5x^{2}=x^{2}-5\left(x+2\right)$$

Use the distributive property to multiply -5 by x+2.

$$7x-16-5x^{2}=x^{2}-5x-10$$

Subtract x^{2} from both sides.

$$7x-16-5x^{2}-x^{2}=-5x-10$$

Combine -5x^{2} and -x^{2} to get -6x^{2}.

$$7x-16-6x^{2}=-5x-10$$

Add 5x to both sides.

$$7x-16-6x^{2}+5x=-10$$

Combine 7x and 5x to get 12x.

$$12x-16-6x^{2}=-10$$

Add 10 to both sides.

$$12x-16-6x^{2}+10=0$$

Add -16 and 10 to get -6.

$$12x-6-6x^{2}=0$$

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 12 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-6\right)\left(-6\right)}}{2\left(-6\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-6\right)\left(-6\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-12±\sqrt{144+24\left(-6\right)}}{2\left(-6\right)}$$

Multiply 24 times -6.

$$x=\frac{-12±\sqrt{144-144}}{2\left(-6\right)}$$

Add 144 to -144.

$$x=\frac{-12±\sqrt{0}}{2\left(-6\right)}$$

Take the square root of 0.

$$x=-\frac{12}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=-\frac{12}{-12}$$

Divide -12 by -12.

$$x=1$$

Steps for Completing the Square

Use the distributive property to multiply 7 by x-3.

$$7x-21-5\left(x^{2}-1\right)=x^{2}-5\left(x+2\right)$$

Use the distributive property to multiply -5 by x^{2}-1.

$$7x-21-5x^{2}+5=x^{2}-5\left(x+2\right)$$

Add -21 and 5 to get -16.

$$7x-16-5x^{2}=x^{2}-5\left(x+2\right)$$

Use the distributive property to multiply -5 by x+2.

$$7x-16-5x^{2}=x^{2}-5x-10$$

Subtract x^{2} from both sides.

$$7x-16-5x^{2}-x^{2}=-5x-10$$

Combine -5x^{2} and -x^{2} to get -6x^{2}.

$$7x-16-6x^{2}=-5x-10$$

Add 5x to both sides.

$$7x-16-6x^{2}+5x=-10$$

Combine 7x and 5x to get 12x.

$$12x-16-6x^{2}=-10$$

Add 16 to both sides.

$$12x-6x^{2}=-10+16$$

Add -10 and 16 to get 6.

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

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.

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

Divide both sides by -6.

$$\frac{-6x^{2}+12x}{-6}=\frac{6}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{12}{-6}x=\frac{6}{-6}$$

Divide 12 by -6.

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

Divide 6 by -6.

$$x^{2}-2x=-1$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=-1+1$$

Add -1 to 1.

$$x^{2}-2x+1=0$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$