Solve 2x^2+18x-20

$$2 x ^ { 2 } + 18 x - 20$$

Factor

$$2\left(x-1\right)\left(x+10\right)$$

Solution Steps

Factor out 2.

$$2\left(x^{2}+9x-10\right)$$

Consider x^{2}+9x-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.

$$a+b=9 ab=1\left(-10\right)=-10$$

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 9.

$$a=-1 b=10$$

Rewrite x^{2}+9x-10 as \left(x^{2}-x\right)+\left(10x-10\right).

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

Factor out x in the first and 10 in the second group.

$$x\left(x-1\right)+10\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+10\right)$$

Rewrite the complete factored expression.

$$2\left(x-1\right)\left(x+10\right)$$

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.

$$2x^{2}+18x-20=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 2\left(-20\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-18±\sqrt{324-8\left(-20\right)}}{2\times 2}$$

Multiply -8 times -20.

$$x=\frac{-18±\sqrt{324+160}}{2\times 2}$$

Add 324 to 160.

$$x=\frac{-18±\sqrt{484}}{2\times 2}$$

Take the square root of 484.

$$x=\frac{-18±22}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-18±22}{4}$$

Now solve the equation x=\frac{-18±22}{4} when ± is plus. Add -18 to 22.

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

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-18±22}{4} when ± is minus. Subtract 22 from -18.

$$x=-\frac{40}{4}$$

Divide -40 by 4.

$$x=-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -10 for x_{2}.

$$2x^{2}+18x-20=2\left(x-1\right)\left(x-\left(-10\right)\right)$$

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

$$2x^{2}+18x-20=2\left(x-1\right)\left(x+10\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 +9x -10 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -9 rs = -10$$

Two numbers r and s sum up to -9 exactly when the average of the two numbers is \frac{1}{2}*-9 = -\frac{9}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{9}{2} - u s = -\frac{9}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(-\frac{9}{2} - u) (-\frac{9}{2} + u) = -10$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{81}{4} - u^2 = -10$$

Simplify the expression by subtracting \frac{81}{4} on both sides

$$-u^2 = -10-\frac{81}{4} = -\frac{121}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{9}{2} - \frac{11}{2} = -10 s = -\frac{9}{2} + \frac{11}{2} = 1$$

Evaluate

$$2\left(x-1\right)\left(x+10\right)$$