Solve -4x^2+30x-100=

$$- 4 x ^ { 2 } + 30 x - 100 =$$

Factor

$$2\left(-2x^{2}+15x-50\right)$$

Solution Steps

Factor out 2. Polynomial -2x^{2}+15x-50 is not factored since it does not have any rational roots.

$$2\left(-2x^{2}+15x-50\right)$$

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.

$$-4x^{2}+30x-100=0$$

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.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-4\right)\left(-100\right)}}{2\left(-4\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-4\right)\left(-100\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-30±\sqrt{900+16\left(-100\right)}}{2\left(-4\right)}$$

Multiply 16 times -100.

$$x=\frac{-30±\sqrt{900-1600}}{2\left(-4\right)}$$

Add 900 to -1600.

$$x=\frac{-30±\sqrt{-700}}{2\left(-4\right)}$$

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

$$-4x^{2}+30x-100$$

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.

$$x ^ 2 -\frac{15}{2}x +25 = 0$$

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

$$r + s = \frac{15}{2} rs = 25$$

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

$$r = \frac{15}{4} - u s = \frac{15}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 25

$$(\frac{15}{4} - u) (\frac{15}{4} + u) = 25$$

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

$$\frac{225}{16} - u^2 = 25$$

Simplify the expression by subtracting \frac{225}{16} on both sides

$$-u^2 = 25-\frac{225}{16} = \frac{175}{16}$$

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

$$u^2 = -\frac{175}{16} u = \pm\sqrt{-\frac{175}{16}} = \pm \frac{\sqrt{175}}{4}i$$

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

$$r =\frac{15}{4} - \frac{\sqrt{175}}{4}i = 3.750 - 3.307i s = \frac{15}{4} + \frac{\sqrt{175}}{4}i = 3.750 + 3.307i$$

Evaluate

$$-4x^{2}+30x-100$$