Solve 4x^2+7x-17=3x^2+12x-3

$$4 x ^ { 2 } + 7 x - 17 = 3 x ^ { 2 } + 12 x - 3$$

Solve for x

$$x=-2$$

$$x=7$$

Steps Using Factoring

Subtract 3x^{2} from both sides.

$$4x^{2}+7x-17-3x^{2}=12x-3$$

Combine 4x^{2} and -3x^{2} to get x^{2}.

$$x^{2}+7x-17=12x-3$$

Subtract 12x from both sides.

$$x^{2}+7x-17-12x=-3$$

Combine 7x and -12x to get -5x.

$$x^{2}-5x-17=-3$$

Add 3 to both sides.

$$x^{2}-5x-17+3=0$$

Add -17 and 3 to get -14.

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

To solve the equation, factor x^{2}-5x-14 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.

$$a+b=-5 ab=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=-7 b=2$$

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

$$\left(x-7\right)\left(x+2\right)$$

To find equation solutions, solve x-7=0 and x+2=0.

$$x=7 x=-2$$

Steps Using Factoring By Grouping

Subtract 3x^{2} from both sides.

$$4x^{2}+7x-17-3x^{2}=12x-3$$

Combine 4x^{2} and -3x^{2} to get x^{2}.

$$x^{2}+7x-17=12x-3$$

Subtract 12x from both sides.

$$x^{2}+7x-17-12x=-3$$

Combine 7x and -12x to get -5x.

$$x^{2}-5x-17=-3$$

Add 3 to both sides.

$$x^{2}-5x-17+3=0$$

Add -17 and 3 to get -14.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-14\right)=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=-7 b=2$$

Rewrite x^{2}-5x-14 as \left(x^{2}-7x\right)+\left(2x-14\right).

$$\left(x^{2}-7x\right)+\left(2x-14\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-7\right)+2\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x+2\right)$$

To find equation solutions, solve x-7=0 and x+2=0.

$$x=7 x=-2$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$4x^{2}+7x-17-3x^{2}=12x-3$$

Combine 4x^{2} and -3x^{2} to get x^{2}.

$$x^{2}+7x-17=12x-3$$

Subtract 12x from both sides.

$$x^{2}+7x-17-12x=-3$$

Combine 7x and -12x to get -5x.

$$x^{2}-5x-17=-3$$

Add 3 to both sides.

$$x^{2}-5x-17+3=0$$

Add -17 and 3 to get -14.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-14\right)}}{2}$$

Multiply -4 times -14.

$$x=\frac{-\left(-5\right)±\sqrt{25+56}}{2}$$

Add 25 to 56.

$$x=\frac{-\left(-5\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-5\right)±9}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±9}{2}$$

Now solve the equation x=\frac{5±9}{2} when ± is plus. Add 5 to 9.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{5±9}{2} when ± is minus. Subtract 9 from 5.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=7 x=-2$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$4x^{2}+7x-17-3x^{2}=12x-3$$

Combine 4x^{2} and -3x^{2} to get x^{2}.

$$x^{2}+7x-17=12x-3$$

Subtract 12x from both sides.

$$x^{2}+7x-17-12x=-3$$

Combine 7x and -12x to get -5x.

$$x^{2}-5x-17=-3$$

Add 17 to both sides.

$$x^{2}-5x=-3+17$$

Add -3 and 17 to get 14.

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

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=14+\frac{25}{4}$$

Add 14 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{81}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{9}{2} x-\frac{5}{2}=-\frac{9}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=7 x=-2$$