Solve 6x^2-15x-20

$$6 x ^ { 2 } - 15 x - 20$$

Factor

$$6\left(x-\left(-\frac{\sqrt{705}}{12}+\frac{5}{4}\right)\right)\left(x-\left(\frac{\sqrt{705}}{12}+\frac{5}{4}\right)\right)$$

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.

$$6x^{2}-15x-20=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 6\left(-20\right)}}{2\times 6}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 6\left(-20\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-15\right)±\sqrt{225-24\left(-20\right)}}{2\times 6}$$

Multiply -24 times -20.

$$x=\frac{-\left(-15\right)±\sqrt{225+480}}{2\times 6}$$

Add 225 to 480.

$$x=\frac{-\left(-15\right)±\sqrt{705}}{2\times 6}$$

The opposite of -15 is 15.

$$x=\frac{15±\sqrt{705}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{15±\sqrt{705}}{12}$$

Now solve the equation x=\frac{15±\sqrt{705}}{12} when ± is plus. Add 15 to \sqrt{705}.

$$x=\frac{\sqrt{705}+15}{12}$$

Divide 15+\sqrt{705} by 12.

$$x=\frac{\sqrt{705}}{12}+\frac{5}{4}$$

Now solve the equation x=\frac{15±\sqrt{705}}{12} when ± is minus. Subtract \sqrt{705} from 15.

$$x=\frac{15-\sqrt{705}}{12}$$

Divide 15-\sqrt{705} by 12.

$$x=-\frac{\sqrt{705}}{12}+\frac{5}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4}+\frac{\sqrt{705}}{12} for x_{1} and \frac{5}{4}-\frac{\sqrt{705}}{12} for x_{2}.

$$6x^{2}-15x-20=6\left(x-\left(\frac{\sqrt{705}}{12}+\frac{5}{4}\right)\right)\left(x-\left(-\frac{\sqrt{705}}{12}+\frac{5}{4}\right)\right)$$

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 6

$$x ^ 2 -\frac{5}{2}x -\frac{10}{3} = 0$$

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

$$r + s = \frac{5}{2} rs = -\frac{10}{3}$$

Two numbers r and s sum up to \frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{2} = \frac{5}{4}. 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>

$$r = \frac{5}{4} - u s = \frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{10}{3}

$$(\frac{5}{4} - u) (\frac{5}{4} + u) = -\frac{10}{3}$$

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

$$\frac{25}{16} - u^2 = -\frac{10}{3}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -\frac{10}{3}-\frac{25}{16} = -\frac{235}{48}$$

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

$$u^2 = \frac{235}{48} u = \pm\sqrt{\frac{235}{48}} = \pm \frac{\sqrt{235}}{\sqrt{48}}$$

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

$$r =\frac{5}{4} - \frac{\sqrt{235}}{\sqrt{48}} = -0.963 s = \frac{5}{4} + \frac{\sqrt{235}}{\sqrt{48}} = 3.463$$

Evaluate

$$6x^{2}-15x-20$$