Solve 3n^2+8n-5

$$3 n ^ { 2 } + 8 n - 5$$

Factor

$$3\left(n-\frac{-\sqrt{31}-4}{3}\right)\left(n-\frac{\sqrt{31}-4}{3}\right)$$

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.

$$3n^{2}+8n-5=0$$

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.

$$n=\frac{-8±\sqrt{8^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square 8.

$$n=\frac{-8±\sqrt{64-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-8±\sqrt{64-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$n=\frac{-8±\sqrt{64+60}}{2\times 3}$$

Add 64 to 60.

$$n=\frac{-8±\sqrt{124}}{2\times 3}$$

Take the square root of 124.

$$n=\frac{-8±2\sqrt{31}}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-8±2\sqrt{31}}{6}$$

Now solve the equation n=\frac{-8±2\sqrt{31}}{6} when ± is plus. Add -8 to 2\sqrt{31}.

$$n=\frac{2\sqrt{31}-8}{6}$$

Divide -8+2\sqrt{31} by 6.

$$n=\frac{\sqrt{31}-4}{3}$$

Now solve the equation n=\frac{-8±2\sqrt{31}}{6} when ± is minus. Subtract 2\sqrt{31} from -8.

$$n=\frac{-2\sqrt{31}-8}{6}$$

Divide -8-2\sqrt{31} by 6.

$$n=\frac{-\sqrt{31}-4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-4+\sqrt{31}}{3} for x_{1} and \frac{-4-\sqrt{31}}{3} for x_{2}.

$$3n^{2}+8n-5=3\left(n-\frac{\sqrt{31}-4}{3}\right)\left(n-\frac{-\sqrt{31}-4}{3}\right)$$

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 3

$$x ^ 2 +\frac{8}{3}x -\frac{5}{3} = 0$$

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

$$r + s = -\frac{8}{3} rs = -\frac{5}{3}$$

Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{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>

$$r = -\frac{4}{3} - u s = -\frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

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

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

$$\frac{16}{9} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = -\frac{5}{3}-\frac{16}{9} = -\frac{31}{9}$$

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

$$u^2 = \frac{31}{9} u = \pm\sqrt{\frac{31}{9}} = \pm \frac{\sqrt{31}}{3}$$

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

$$r =-\frac{4}{3} - \frac{\sqrt{31}}{3} = -3.189 s = -\frac{4}{3} + \frac{\sqrt{31}}{3} = 0.523$$

Evaluate

$$3n^{2}+8n-5$$