Solve 3x^2+15x-30

$$3 x ^ { 2 } + 15 x - 30$$

Factor

$$3\left(x-\frac{-\sqrt{65}-5}{2}\right)\left(x-\frac{\sqrt{65}-5}{2}\right)$$

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.

$$3x^{2}+15x-30=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\times 3\left(-30\right)}}{2\times 3}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 3\left(-30\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-15±\sqrt{225-12\left(-30\right)}}{2\times 3}$$

Multiply -12 times -30.

$$x=\frac{-15±\sqrt{225+360}}{2\times 3}$$

Add 225 to 360.

$$x=\frac{-15±\sqrt{585}}{2\times 3}$$

Take the square root of 585.

$$x=\frac{-15±3\sqrt{65}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-15±3\sqrt{65}}{6}$$

Now solve the equation x=\frac{-15±3\sqrt{65}}{6} when ± is plus. Add -15 to 3\sqrt{65}.

$$x=\frac{3\sqrt{65}-15}{6}$$

Divide -15+3\sqrt{65} by 6.

$$x=\frac{\sqrt{65}-5}{2}$$

Now solve the equation x=\frac{-15±3\sqrt{65}}{6} when ± is minus. Subtract 3\sqrt{65} from -15.

$$x=\frac{-3\sqrt{65}-15}{6}$$

Divide -15-3\sqrt{65} by 6.

$$x=\frac{-\sqrt{65}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{65}}{2} for x_{1} and \frac{-5-\sqrt{65}}{2} for x_{2}.

$$3x^{2}+15x-30=3\left(x-\frac{\sqrt{65}-5}{2}\right)\left(x-\frac{-\sqrt{65}-5}{2}\right)$$

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

$$x ^ 2 +5x -10 = 0$$

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

$$r + s = -5 rs = -10$$

Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{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>

$$r = -\frac{5}{2} - u s = -\frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -10$$

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

$$\frac{25}{4} - u^2 = -10$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -10-\frac{25}{4} = -\frac{65}{4}$$

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

$$u^2 = \frac{65}{4} u = \pm\sqrt{\frac{65}{4}} = \pm \frac{\sqrt{65}}{2}$$

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

$$r =-\frac{5}{2} - \frac{\sqrt{65}}{2} = -6.531 s = -\frac{5}{2} + \frac{\sqrt{65}}{2} = 1.531$$

Evaluate

$$3\left(x^{2}+5x-10\right)$$