Solve -16x^2+92x

$$-16 { x }^{ 2 } +92x$$

Factor

$$4x\left(23-4x\right)$$

Solution Steps

Factor out 4.

$$4\left(-4x^{2}+23x\right)$$

Consider -4x^{2}+23x. Factor out x.

$$x\left(-4x+23\right)$$

Rewrite the complete factored expression.

$$4x\left(-4x+23\right)$$

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.

$$-16x^{2}+92x=0$$

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.

$$x=\frac{-92±\sqrt{92^{2}}}{2\left(-16\right)}$$

Take the square root of 92^{2}.

$$x=\frac{-92±92}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-92±92}{-32}$$

Now solve the equation x=\frac{-92±92}{-32} when ± is plus. Add -92 to 92.

$$x=\frac{0}{-32}$$

Divide 0 by -32.

$$x=0$$

Now solve the equation x=\frac{-92±92}{-32} when ± is minus. Subtract 92 from -92.

$$x=-\frac{184}{-32}$$

Reduce the fraction \frac{-184}{-32} to lowest terms by extracting and canceling out 8.

$$x=\frac{23}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and \frac{23}{4} for x_{2}.

$$-16x^{2}+92x=-16x\left(x-\frac{23}{4}\right)$$

Subtract \frac{23}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-16x^{2}+92x=-16x\times \frac{-4x+23}{-4}$$

Cancel out 4, the greatest common factor in -16 and -4.

$$-16x^{2}+92x=4x\left(-4x+23\right)$$

Evaluate

$$4x\left(23-4x\right)$$