Solve 15x^2+64x-9

$$15 x ^ { 2 } + 64 x - 9$$

Factor

$$15\left(x-\frac{-\sqrt{1159}-32}{15}\right)\left(x-\frac{\sqrt{1159}-32}{15}\right)$$

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.

$$15x^{2}+64x-9=0$$

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.

$$x=\frac{-64±\sqrt{64^{2}-4\times 15\left(-9\right)}}{2\times 15}$$

Square 64.

$$x=\frac{-64±\sqrt{4096-4\times 15\left(-9\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-64±\sqrt{4096-60\left(-9\right)}}{2\times 15}$$

Multiply -60 times -9.

$$x=\frac{-64±\sqrt{4096+540}}{2\times 15}$$

Add 4096 to 540.

$$x=\frac{-64±\sqrt{4636}}{2\times 15}$$

Take the square root of 4636.

$$x=\frac{-64±2\sqrt{1159}}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-64±2\sqrt{1159}}{30}$$

Now solve the equation x=\frac{-64±2\sqrt{1159}}{30} when ± is plus. Add -64 to 2\sqrt{1159}.

$$x=\frac{2\sqrt{1159}-64}{30}$$

Divide -64+2\sqrt{1159} by 30.

$$x=\frac{\sqrt{1159}-32}{15}$$

Now solve the equation x=\frac{-64±2\sqrt{1159}}{30} when ± is minus. Subtract 2\sqrt{1159} from -64.

$$x=\frac{-2\sqrt{1159}-64}{30}$$

Divide -64-2\sqrt{1159} by 30.

$$x=\frac{-\sqrt{1159}-32}{15}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-32+\sqrt{1159}}{15} for x_{1} and \frac{-32-\sqrt{1159}}{15} for x_{2}.

$$15x^{2}+64x-9=15\left(x-\frac{\sqrt{1159}-32}{15}\right)\left(x-\frac{-\sqrt{1159}-32}{15}\right)$$

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 15

$$x ^ 2 +\frac{64}{15}x -\frac{3}{5} = 0$$

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

$$r + s = -\frac{64}{15} rs = -\frac{3}{5}$$

Two numbers r and s sum up to -\frac{64}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{64}{15} = -\frac{32}{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>

$$r = -\frac{32}{15} - u s = -\frac{32}{15} + u$$

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

$$(-\frac{32}{15} - u) (-\frac{32}{15} + u) = -\frac{3}{5}$$

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

$$\frac{1024}{225} - u^2 = -\frac{3}{5}$$

Simplify the expression by subtracting \frac{1024}{225} on both sides

$$-u^2 = -\frac{3}{5}-\frac{1024}{225} = -\frac{1159}{225}$$

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

$$u^2 = \frac{1159}{225} u = \pm\sqrt{\frac{1159}{225}} = \pm \frac{\sqrt{1159}}{15}$$

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

$$r =-\frac{32}{15} - \frac{\sqrt{1159}}{15} = -4.403 s = -\frac{32}{15} + \frac{\sqrt{1159}}{15} = 0.136$$

Evaluate

$$15x^{2}+64x-9$$