Solve [3x(5+x)]+[2*(-x+3)]=2x+[5*(x+3)]

$$[ 3 x ( 5 + x ) ] + [ 2 \times ( - x + 3 ) ] = 2 x + [ 5 \times ( x + 3 ) ]$$

Solve for x

$$x=-3$$

$$x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 5+x.

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

Use the distributive property to multiply 2 by -x+3.

$$15x+3x^{2}+2\left(-x\right)+6=2x+5\left(x+3\right)$$

Use the distributive property to multiply 5 by x+3.

$$15x+3x^{2}+2\left(-x\right)+6=2x+5x+15$$

Combine 2x and 5x to get 7x.

$$15x+3x^{2}+2\left(-x\right)+6=7x+15$$

Subtract 7x from both sides.

$$15x+3x^{2}+2\left(-x\right)+6-7x=15$$

Combine 15x and -7x to get 8x.

$$8x+3x^{2}+2\left(-x\right)+6=15$$

Subtract 15 from both sides.

$$8x+3x^{2}+2\left(-x\right)+6-15=0$$

Subtract 15 from 6 to get -9.

$$8x+3x^{2}+2\left(-x\right)-9=0$$

Multiply 2 and -1 to get -2.

$$8x+3x^{2}-2x-9=0$$

Combine 8x and -2x to get 6x.

$$6x+3x^{2}-9=0$$

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.

$$3x^{2}+6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 3\left(-9\right)}}{2\times 3}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 3\left(-9\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-6±\sqrt{36-12\left(-9\right)}}{2\times 3}$$

Multiply -12 times -9.

$$x=\frac{-6±\sqrt{36+108}}{2\times 3}$$

Add 36 to 108.

$$x=\frac{-6±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{-6±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-6±12}{6}$$

Now solve the equation x=\frac{-6±12}{6} when ± is plus. Add -6 to 12.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{-6±12}{6} when ± is minus. Subtract 12 from -6.

$$x=-\frac{18}{6}$$

Divide -18 by 6.

$$x=-3$$

The equation is now solved.

$$x=1 x=-3$$

Steps for Completing the Square

Use the distributive property to multiply 3x by 5+x.

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

Use the distributive property to multiply 2 by -x+3.

$$15x+3x^{2}+2\left(-x\right)+6=2x+5\left(x+3\right)$$

Use the distributive property to multiply 5 by x+3.

$$15x+3x^{2}+2\left(-x\right)+6=2x+5x+15$$

Combine 2x and 5x to get 7x.

$$15x+3x^{2}+2\left(-x\right)+6=7x+15$$

Subtract 7x from both sides.

$$15x+3x^{2}+2\left(-x\right)+6-7x=15$$

Combine 15x and -7x to get 8x.

$$8x+3x^{2}+2\left(-x\right)+6=15$$

Subtract 6 from both sides.

$$8x+3x^{2}+2\left(-x\right)=15-6$$

Subtract 6 from 15 to get 9.

$$8x+3x^{2}+2\left(-x\right)=9$$

Multiply 2 and -1 to get -2.

$$8x+3x^{2}-2x=9$$

Combine 8x and -2x to get 6x.

$$6x+3x^{2}=9$$

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.

$$3x^{2}+6x=9$$

Divide both sides by 3.

$$\frac{3x^{2}+6x}{3}=\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{6}{3}x=\frac{9}{3}$$

Divide 6 by 3.

$$x^{2}+2x=\frac{9}{3}$$

Divide 9 by 3.

$$x^{2}+2x=3$$

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.

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

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}.

$$\left(x+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$