Solve 12x^2+14x-10

$$12 { x }^{ 2 } +14x-10$$

Factor

$$2\left(2x-1\right)\left(3x+5\right)$$

Solution Steps

Factor out 2.

$$2\left(6x^{2}+7x-5\right)$$

Consider 6x^{2}+7x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=7 ab=6\left(-5\right)=-30$$

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 7.

$$a=-3 b=10$$

Rewrite 6x^{2}+7x-5 as \left(6x^{2}-3x\right)+\left(10x-5\right).

$$\left(6x^{2}-3x\right)+\left(10x-5\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(2x-1\right)+5\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(3x+5\right)$$

Rewrite the complete factored expression.

$$2\left(2x-1\right)\left(3x+5\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$12x^{2}+14x-10=0$$

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.

$$x=\frac{-14±\sqrt{14^{2}-4\times 12\left(-10\right)}}{2\times 12}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 12\left(-10\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-14±\sqrt{196-48\left(-10\right)}}{2\times 12}$$

Multiply -48 times -10.

$$x=\frac{-14±\sqrt{196+480}}{2\times 12}$$

Add 196 to 480.

$$x=\frac{-14±\sqrt{676}}{2\times 12}$$

Take the square root of 676.

$$x=\frac{-14±26}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-14±26}{24}$$

Now solve the equation x=\frac{-14±26}{24} when ± is plus. Add -14 to 26.

$$x=\frac{12}{24}$$

Reduce the fraction \frac{12}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-14±26}{24} when ± is minus. Subtract 26 from -14.

$$x=-\frac{40}{24}$$

Reduce the fraction \frac{-40}{24} to lowest terms by extracting and canceling out 8.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2} for x_{1} and -\frac{5}{3} for x_{2}.

$$12x^{2}+14x-10=12\left(x-\frac{1}{2}\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$12x^{2}+14x-10=12\left(x-\frac{1}{2}\right)\left(x+\frac{5}{3}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+14x-10=12\times \frac{2x-1}{2}\left(x+\frac{5}{3}\right)$$

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

$$12x^{2}+14x-10=12\times \frac{2x-1}{2}\times \frac{3x+5}{3}$$

Multiply \frac{2x-1}{2} times \frac{3x+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+14x-10=12\times \frac{\left(2x-1\right)\left(3x+5\right)}{2\times 3}$$

Multiply 2 times 3.

$$12x^{2}+14x-10=12\times \frac{\left(2x-1\right)\left(3x+5\right)}{6}$$

Cancel out 6, the greatest common factor in 12 and 6.

$$12x^{2}+14x-10=2\left(2x-1\right)\left(3x+5\right)$$

Evaluate

$$12x^{2}+14x-10$$