Solve 3x^2+13x-28=2x^2+2

$$3 x ^ { 2 } + 13 x - 28 = 2 x ^ { 2 } + 2$$

Solve for x

$$x=-15$$

$$x=2$$

Steps Using Factoring

Subtract 2x^{2} from both sides.

$$3x^{2}+13x-28-2x^{2}=2$$

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

$$x^{2}+13x-28=2$$

Subtract 2 from both sides.

$$x^{2}+13x-28-2=0$$

Subtract 2 from -28 to get -30.

$$x^{2}+13x-30=0$$

To solve the equation, factor x^{2}+13x-30 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=13 ab=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 13.

$$a=-2 b=15$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-2\right)\left(x+15\right)$$

To find equation solutions, solve x-2=0 and x+15=0.

$$x=2 x=-15$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$3x^{2}+13x-28-2x^{2}=2$$

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

$$x^{2}+13x-28=2$$

Subtract 2 from both sides.

$$x^{2}+13x-28-2=0$$

Subtract 2 from -28 to get -30.

$$x^{2}+13x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=13 ab=1\left(-30\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 13.

$$a=-2 b=15$$

Rewrite x^{2}+13x-30 as \left(x^{2}-2x\right)+\left(15x-30\right).

$$\left(x^{2}-2x\right)+\left(15x-30\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(x-2\right)+15\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+15\right)$$

To find equation solutions, solve x-2=0 and x+15=0.

$$x=2 x=-15$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$3x^{2}+13x-28-2x^{2}=2$$

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

$$x^{2}+13x-28=2$$

Subtract 2 from both sides.

$$x^{2}+13x-28-2=0$$

Subtract 2 from -28 to get -30.

$$x^{2}+13x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 13 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-30\right)}}{2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$x=\frac{-13±\sqrt{169+120}}{2}$$

Add 169 to 120.

$$x=\frac{-13±\sqrt{289}}{2}$$

Take the square root of 289.

$$x=\frac{-13±17}{2}$$

Now solve the equation x=\frac{-13±17}{2} when ± is plus. Add -13 to 17.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{-13±17}{2} when ± is minus. Subtract 17 from -13.

$$x=-\frac{30}{2}$$

Divide -30 by 2.

$$x=-15$$

The equation is now solved.

$$x=2 x=-15$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$3x^{2}+13x-28-2x^{2}=2$$

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

$$x^{2}+13x-28=2$$

Add 28 to both sides.

$$x^{2}+13x=2+28$$

Add 2 and 28 to get 30.

$$x^{2}+13x=30$$

Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+13x+\left(\frac{13}{2}\right)^{2}=30+\left(\frac{13}{2}\right)^{2}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+13x+\frac{169}{4}=30+\frac{169}{4}$$

Add 30 to \frac{169}{4}.

$$x^{2}+13x+\frac{169}{4}=\frac{289}{4}$$

Factor x^{2}+13x+\frac{169}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x+\frac{13}{2}=\frac{17}{2} x+\frac{13}{2}=-\frac{17}{2}$$

Subtract \frac{13}{2} from both sides of the equation.

$$x=2 x=-15$$