Solve 20z^2+24z-9

$$20 z ^ { 2 } + 24 z - 9$$

Factor

$$\left(10z-3\right)\left(2z+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20z^{2}+az+bz-9. To find a and b, set up a system to be solved.

$$a+b=24 ab=20\left(-9\right)=-180$$

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 -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 24.

$$a=-6 b=30$$

Rewrite 20z^{2}+24z-9 as \left(20z^{2}-6z\right)+\left(30z-9\right).

$$\left(20z^{2}-6z\right)+\left(30z-9\right)$$

Factor out 2z in the first and 3 in the second group.

$$2z\left(10z-3\right)+3\left(10z-3\right)$$

Factor out common term 10z-3 by using distributive property.

$$\left(10z-3\right)\left(2z+3\right)$$

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.

$$20z^{2}+24z-9=0$$

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.

$$z=\frac{-24±\sqrt{24^{2}-4\times 20\left(-9\right)}}{2\times 20}$$

Square 24.

$$z=\frac{-24±\sqrt{576-4\times 20\left(-9\right)}}{2\times 20}$$

Multiply -4 times 20.

$$z=\frac{-24±\sqrt{576-80\left(-9\right)}}{2\times 20}$$

Multiply -80 times -9.

$$z=\frac{-24±\sqrt{576+720}}{2\times 20}$$

Add 576 to 720.

$$z=\frac{-24±\sqrt{1296}}{2\times 20}$$

Take the square root of 1296.

$$z=\frac{-24±36}{2\times 20}$$

Multiply 2 times 20.

$$z=\frac{-24±36}{40}$$

Now solve the equation z=\frac{-24±36}{40} when ± is plus. Add -24 to 36.

$$z=\frac{12}{40}$$

Reduce the fraction \frac{12}{40} to lowest terms by extracting and canceling out 4.

$$z=\frac{3}{10}$$

Now solve the equation z=\frac{-24±36}{40} when ± is minus. Subtract 36 from -24.

$$z=-\frac{60}{40}$$

Reduce the fraction \frac{-60}{40} to lowest terms by extracting and canceling out 20.

$$z=-\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{10} for x_{1} and -\frac{3}{2} for x_{2}.

$$20z^{2}+24z-9=20\left(z-\frac{3}{10}\right)\left(z-\left(-\frac{3}{2}\right)\right)$$

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

$$20z^{2}+24z-9=20\left(z-\frac{3}{10}\right)\left(z+\frac{3}{2}\right)$$

Subtract \frac{3}{10} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20z^{2}+24z-9=20\times \frac{10z-3}{10}\left(z+\frac{3}{2}\right)$$

Add \frac{3}{2} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20z^{2}+24z-9=20\times \frac{10z-3}{10}\times \frac{2z+3}{2}$$

Multiply \frac{10z-3}{10} times \frac{2z+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20z^{2}+24z-9=20\times \frac{\left(10z-3\right)\left(2z+3\right)}{10\times 2}$$

Multiply 10 times 2.

$$20z^{2}+24z-9=20\times \frac{\left(10z-3\right)\left(2z+3\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20z^{2}+24z-9=\left(10z-3\right)\left(2z+3\right)$$

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 20

$$x ^ 2 +\frac{6}{5}x -\frac{9}{20} = 0$$

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

$$r + s = -\frac{6}{5} rs = -\frac{9}{20}$$

Two numbers r and s sum up to -\frac{6}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{6}{5} = -\frac{3}{5}. 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>

$$r = -\frac{3}{5} - u s = -\frac{3}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{20}

$$(-\frac{3}{5} - u) (-\frac{3}{5} + u) = -\frac{9}{20}$$

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

$$\frac{9}{25} - u^2 = -\frac{9}{20}$$

Simplify the expression by subtracting \frac{9}{25} on both sides

$$-u^2 = -\frac{9}{20}-\frac{9}{25} = -\frac{81}{100}$$

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

$$u^2 = \frac{81}{100} u = \pm\sqrt{\frac{81}{100}} = \pm \frac{9}{10}$$

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

$$r =-\frac{3}{5} - \frac{9}{10} = -1.500 s = -\frac{3}{5} + \frac{9}{10} = 0.300$$

Evaluate

$$\left(10z-3\right)\left(2z+3\right)$$