Solve 5x^2+20x-6

$$5 x ^ { 2 } + 20 x - 6$$

Factor

$$5\left(x-\left(-\frac{\sqrt{130}}{5}-2\right)\right)\left(x-\left(\frac{\sqrt{130}}{5}-2\right)\right)$$

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.

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

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 5\left(-6\right)}}{2\times 5}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 5\left(-6\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-20±\sqrt{400-20\left(-6\right)}}{2\times 5}$$

Multiply -20 times -6.

$$x=\frac{-20±\sqrt{400+120}}{2\times 5}$$

Add 400 to 120.

$$x=\frac{-20±\sqrt{520}}{2\times 5}$$

Take the square root of 520.

$$x=\frac{-20±2\sqrt{130}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-20±2\sqrt{130}}{10}$$

Now solve the equation x=\frac{-20±2\sqrt{130}}{10} when ± is plus. Add -20 to 2\sqrt{130}.

$$x=\frac{2\sqrt{130}-20}{10}$$

Divide -20+2\sqrt{130} by 10.

$$x=\frac{\sqrt{130}}{5}-2$$

Now solve the equation x=\frac{-20±2\sqrt{130}}{10} when ± is minus. Subtract 2\sqrt{130} from -20.

$$x=\frac{-2\sqrt{130}-20}{10}$$

Divide -20-2\sqrt{130} by 10.

$$x=-\frac{\sqrt{130}}{5}-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2+\frac{\sqrt{130}}{5} for x_{1} and -2-\frac{\sqrt{130}}{5} for x_{2}.

$$5x^{2}+20x-6=5\left(x-\left(\frac{\sqrt{130}}{5}-2\right)\right)\left(x-\left(-\frac{\sqrt{130}}{5}-2\right)\right)$$

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 5

$$x ^ 2 +4x -\frac{6}{5} = 0$$

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

$$r + s = -4 rs = -\frac{6}{5}$$

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

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{6}{5}

$$(-2 - u) (-2 + u) = -\frac{6}{5}$$

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

$$4 - u^2 = -\frac{6}{5}$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -\frac{6}{5}-4 = -\frac{26}{5}$$

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

$$u^2 = \frac{26}{5} u = \pm\sqrt{\frac{26}{5}} = \pm \frac{\sqrt{26}}{\sqrt{5}}$$

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

$$r =-2 - \frac{\sqrt{26}}{\sqrt{5}} = -4.280 s = -2 + \frac{\sqrt{26}}{\sqrt{5}} = 0.280$$

Evaluate

$$5x^{2}+20x-6$$