Solve 30x^2+4x-8

$$30 x ^ { 2 } + 4 x - 8$$

Factor

$$30\left(x-\frac{-\sqrt{61}-1}{15}\right)\left(x-\frac{\sqrt{61}-1}{15}\right)$$

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.

$$30x^{2}+4x-8=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\times 30\left(-8\right)}}{2\times 30}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 30\left(-8\right)}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-4±\sqrt{16-120\left(-8\right)}}{2\times 30}$$

Multiply -120 times -8.

$$x=\frac{-4±\sqrt{16+960}}{2\times 30}$$

Add 16 to 960.

$$x=\frac{-4±\sqrt{976}}{2\times 30}$$

Take the square root of 976.

$$x=\frac{-4±4\sqrt{61}}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{-4±4\sqrt{61}}{60}$$

Now solve the equation x=\frac{-4±4\sqrt{61}}{60} when ± is plus. Add -4 to 4\sqrt{61}.

$$x=\frac{4\sqrt{61}-4}{60}$$

Divide -4+4\sqrt{61} by 60.

$$x=\frac{\sqrt{61}-1}{15}$$

Now solve the equation x=\frac{-4±4\sqrt{61}}{60} when ± is minus. Subtract 4\sqrt{61} from -4.

$$x=\frac{-4\sqrt{61}-4}{60}$$

Divide -4-4\sqrt{61} by 60.

$$x=\frac{-\sqrt{61}-1}{15}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{61}}{15} for x_{1} and \frac{-1-\sqrt{61}}{15} for x_{2}.

$$30x^{2}+4x-8=30\left(x-\frac{\sqrt{61}-1}{15}\right)\left(x-\frac{-\sqrt{61}-1}{15}\right)$$

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 30

$$x ^ 2 +\frac{2}{15}x -\frac{4}{15} = 0$$

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

$$r + s = -\frac{2}{15} rs = -\frac{4}{15}$$

Two numbers r and s sum up to -\frac{2}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{15} = -\frac{1}{15}. 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>

$$r = -\frac{1}{15} - u s = -\frac{1}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{15}

$$(-\frac{1}{15} - u) (-\frac{1}{15} + u) = -\frac{4}{15}$$

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

$$\frac{1}{225} - u^2 = -\frac{4}{15}$$

Simplify the expression by subtracting \frac{1}{225} on both sides

$$-u^2 = -\frac{4}{15}-\frac{1}{225} = -\frac{61}{225}$$

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

$$u^2 = \frac{61}{225} u = \pm\sqrt{\frac{61}{225}} = \pm \frac{\sqrt{61}}{15}$$

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

$$r =-\frac{1}{15} - \frac{\sqrt{61}}{15} = -0.587 s = -\frac{1}{15} + \frac{\sqrt{61}}{15} = 0.454$$

Evaluate

$$30x^{2}+4x-8$$