Solve 15x^2-20x+10=

$$15 x ^ { 2 } - 20 x + 10 =$$

Factor

$$5\left(3x^{2}-4x+2\right)$$

Solution Steps

Factor out 5. Polynomial 3x^{2}-4x+2 is not factored since it does not have any rational roots.

$$5\left(3x^{2}-4x+2\right)$$

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.

$$15x^{2}-20x+10=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 15\times 10}}{2\times 15}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 15\times 10}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-20\right)±\sqrt{400-60\times 10}}{2\times 15}$$

Multiply -60 times 10.

$$x=\frac{-\left(-20\right)±\sqrt{400-600}}{2\times 15}$$

Add 400 to -600.

$$x=\frac{-\left(-20\right)±\sqrt{-200}}{2\times 15}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$15x^{2}-20x+10$$

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

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

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

$$r + s = \frac{4}{3} rs = \frac{2}{3}$$

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

$$r = \frac{2}{3} - u s = \frac{2}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}

$$(\frac{2}{3} - u) (\frac{2}{3} + u) = \frac{2}{3}$$

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

$$\frac{4}{9} - u^2 = \frac{2}{3}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = \frac{2}{3}-\frac{4}{9} = \frac{2}{9}$$

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

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

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

$$r =\frac{2}{3} - \frac{\sqrt{2}}{3}i = 0.667 - 0.471i s = \frac{2}{3} + \frac{\sqrt{2}}{3}i = 0.667 + 0.471i$$

Evaluate

$$15x^{2}-20x+10$$