Solve 12x^2+16x-5

$$12 x ^ { 2 } + 16 x - 5$$

Factor

$$12\left(x-\left(-\frac{\sqrt{31}}{6}-\frac{2}{3}\right)\right)\left(x-\left(\frac{\sqrt{31}}{6}-\frac{2}{3}\right)\right)$$

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.

$$12x^{2}+16x-5=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 12\left(-5\right)}}{2\times 12}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 12\left(-5\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-16±\sqrt{256-48\left(-5\right)}}{2\times 12}$$

Multiply -48 times -5.

$$x=\frac{-16±\sqrt{256+240}}{2\times 12}$$

Add 256 to 240.

$$x=\frac{-16±\sqrt{496}}{2\times 12}$$

Take the square root of 496.

$$x=\frac{-16±4\sqrt{31}}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-16±4\sqrt{31}}{24}$$

Now solve the equation x=\frac{-16±4\sqrt{31}}{24} when ± is plus. Add -16 to 4\sqrt{31}.

$$x=\frac{4\sqrt{31}-16}{24}$$

Divide -16+4\sqrt{31} by 24.

$$x=\frac{\sqrt{31}}{6}-\frac{2}{3}$$

Now solve the equation x=\frac{-16±4\sqrt{31}}{24} when ± is minus. Subtract 4\sqrt{31} from -16.

$$x=\frac{-4\sqrt{31}-16}{24}$$

Divide -16-4\sqrt{31} by 24.

$$x=-\frac{\sqrt{31}}{6}-\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3}+\frac{\sqrt{31}}{6} for x_{1} and -\frac{2}{3}-\frac{\sqrt{31}}{6} for x_{2}.

$$12x^{2}+16x-5=12\left(x-\left(\frac{\sqrt{31}}{6}-\frac{2}{3}\right)\right)\left(x-\left(-\frac{\sqrt{31}}{6}-\frac{2}{3}\right)\right)$$

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 12

$$x ^ 2 +\frac{4}{3}x -\frac{5}{12} = 0$$

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

$$r + s = -\frac{4}{3} rs = -\frac{5}{12}$$

Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{3}. 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>

$$r = -\frac{2}{3} - u s = -\frac{2}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{12}

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = -\frac{5}{12}$$

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

$$\frac{4}{9} - u^2 = -\frac{5}{12}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = -\frac{5}{12}-\frac{4}{9} = -\frac{31}{36}$$

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

$$u^2 = \frac{31}{36} u = \pm\sqrt{\frac{31}{36}} = \pm \frac{\sqrt{31}}{6}$$

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

$$r =-\frac{2}{3} - \frac{\sqrt{31}}{6} = -1.595 s = -\frac{2}{3} + \frac{\sqrt{31}}{6} = 0.261$$

Evaluate

$$12x^{2}+16x-5$$