Solve 6x+2x^2-20

$$6 x + 2 x ^ { 2 } - 20$$

Factor

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

Solution Steps

Factor out 2.

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

Consider 3x+x^{2}-10. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+3x-10$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-10\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$2x^{2}+6x-20=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-20\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\left(-20\right)}}{2\times 2}$$

Multiply -8 times -20.

$$x=\frac{-6±\sqrt{36+160}}{2\times 2}$$

Add 36 to 160.

$$x=\frac{-6±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$x=\frac{-6±14}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±14}{4}$$

Now solve the equation x=\frac{-6±14}{4} when ± is plus. Add -6 to 14.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{-6±14}{4} when ± is minus. Subtract 14 from -6.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

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

$$2x^{2}+6x-20=2\left(x-2\right)\left(x-\left(-5\right)\right)$$

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

$$2x^{2}+6x-20=2\left(x-2\right)\left(x+5\right)$$

Evaluate

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