Solve 9m^2+28m-36

$$9 m ^ { 2 } + 28 m - 36$$

Factor

$$9\left(m-\frac{-2\sqrt{130}-14}{9}\right)\left(m-\frac{2\sqrt{130}-14}{9}\right)$$

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.

$$9m^{2}+28m-36=0$$

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.

$$m=\frac{-28±\sqrt{28^{2}-4\times 9\left(-36\right)}}{2\times 9}$$

Square 28.

$$m=\frac{-28±\sqrt{784-4\times 9\left(-36\right)}}{2\times 9}$$

Multiply -4 times 9.

$$m=\frac{-28±\sqrt{784-36\left(-36\right)}}{2\times 9}$$

Multiply -36 times -36.

$$m=\frac{-28±\sqrt{784+1296}}{2\times 9}$$

Add 784 to 1296.

$$m=\frac{-28±\sqrt{2080}}{2\times 9}$$

Take the square root of 2080.

$$m=\frac{-28±4\sqrt{130}}{2\times 9}$$

Multiply 2 times 9.

$$m=\frac{-28±4\sqrt{130}}{18}$$

Now solve the equation m=\frac{-28±4\sqrt{130}}{18} when ± is plus. Add -28 to 4\sqrt{130}.

$$m=\frac{4\sqrt{130}-28}{18}$$

Divide -28+4\sqrt{130} by 18.

$$m=\frac{2\sqrt{130}-14}{9}$$

Now solve the equation m=\frac{-28±4\sqrt{130}}{18} when ± is minus. Subtract 4\sqrt{130} from -28.

$$m=\frac{-4\sqrt{130}-28}{18}$$

Divide -28-4\sqrt{130} by 18.

$$m=\frac{-2\sqrt{130}-14}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-14+2\sqrt{130}}{9} for x_{1} and \frac{-14-2\sqrt{130}}{9} for x_{2}.

$$9m^{2}+28m-36=9\left(m-\frac{2\sqrt{130}-14}{9}\right)\left(m-\frac{-2\sqrt{130}-14}{9}\right)$$

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 9

$$x ^ 2 +\frac{28}{9}x -4 = 0$$

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

$$r + s = -\frac{28}{9} rs = -4$$

Two numbers r and s sum up to -\frac{28}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{28}{9} = -\frac{14}{9}. 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>

$$r = -\frac{14}{9} - u s = -\frac{14}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(-\frac{14}{9} - u) (-\frac{14}{9} + u) = -4$$

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

$$\frac{196}{81} - u^2 = -4$$

Simplify the expression by subtracting \frac{196}{81} on both sides

$$-u^2 = -4-\frac{196}{81} = -\frac{520}{81}$$

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

$$u^2 = \frac{520}{81} u = \pm\sqrt{\frac{520}{81}} = \pm \frac{\sqrt{520}}{9}$$

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

$$r =-\frac{14}{9} - \frac{\sqrt{520}}{9} = -4.089 s = -\frac{14}{9} + \frac{\sqrt{520}}{9} = 0.978$$

Evaluate

$$9m^{2}+28m-36$$