Solve 8x^2+4x-3

$$8 x ^ { 2 } + 4 x - 3$$

Factor

$$8\left(x-\frac{-\sqrt{7}-1}{4}\right)\left(x-\frac{\sqrt{7}-1}{4}\right)$$

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.

$$8x^{2}+4x-3=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 8\left(-3\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-4±\sqrt{16-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$x=\frac{-4±\sqrt{16+96}}{2\times 8}$$

Add 16 to 96.

$$x=\frac{-4±\sqrt{112}}{2\times 8}$$

Take the square root of 112.

$$x=\frac{-4±4\sqrt{7}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-4±4\sqrt{7}}{16}$$

Now solve the equation x=\frac{-4±4\sqrt{7}}{16} when ± is plus. Add -4 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}-4}{16}$$

Divide -4+4\sqrt{7} by 16.

$$x=\frac{\sqrt{7}-1}{4}$$

Now solve the equation x=\frac{-4±4\sqrt{7}}{16} when ± is minus. Subtract 4\sqrt{7} from -4.

$$x=\frac{-4\sqrt{7}-4}{16}$$

Divide -4-4\sqrt{7} by 16.

$$x=\frac{-\sqrt{7}-1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{7}}{4} for x_{1} and \frac{-1-\sqrt{7}}{4} for x_{2}.

$$8x^{2}+4x-3=8\left(x-\frac{\sqrt{7}-1}{4}\right)\left(x-\frac{-\sqrt{7}-1}{4}\right)$$

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 8

$$x ^ 2 +\frac{1}{2}x -\frac{3}{8} = 0$$

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

$$r + s = -\frac{1}{2} rs = -\frac{3}{8}$$

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

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{8}

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -\frac{3}{8}$$

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

$$\frac{1}{16} - u^2 = -\frac{3}{8}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{3}{8}-\frac{1}{16} = -\frac{7}{16}$$

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

$$u^2 = \frac{7}{16} u = \pm\sqrt{\frac{7}{16}} = \pm \frac{\sqrt{7}}{4}$$

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

$$r =-\frac{1}{4} - \frac{\sqrt{7}}{4} = -0.911 s = -\frac{1}{4} + \frac{\sqrt{7}}{4} = 0.411$$

Evaluate

$$8x^{2}+4x-3$$