Solve -50=-30x+4x^2

$$- 50 = - 30 x + 4 x ^ { 2 }$$

Solve for x

$$x=5$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$-30x+4x^{2}=-50$$

Add 50 to both sides.

$$-30x+4x^{2}+50=0$$

Divide both sides by 2.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-15 ab=2\times 25=50$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 50.

$$-1,-50 -2,-25 -5,-10$$

Calculate the sum for each pair.

$$-1-50=-51 -2-25=-27 -5-10=-15$$

The solution is the pair that gives sum -15.

$$a=-10 b=-5$$

Rewrite 2x^{2}-15x+25 as \left(2x^{2}-10x\right)+\left(-5x+25\right).

$$\left(2x^{2}-10x\right)+\left(-5x+25\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-5\right)-5\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(2x-5\right)$$

To find equation solutions, solve x-5=0 and 2x-5=0.

$$x=5 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-30x+4x^{2}=-50$$

Add 50 to both sides.

$$-30x+4x^{2}+50=0$$

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.

$$4x^{2}-30x+50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -30 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 4\times 50}}{2\times 4}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 4\times 50}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-30\right)±\sqrt{900-16\times 50}}{2\times 4}$$

Multiply -16 times 50.

$$x=\frac{-\left(-30\right)±\sqrt{900-800}}{2\times 4}$$

Add 900 to -800.

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

Take the square root of 100.

$$x=\frac{-\left(-30\right)±10}{2\times 4}$$

The opposite of -30 is 30.

$$x=\frac{30±10}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{30±10}{8}$$

Now solve the equation x=\frac{30±10}{8} when ± is plus. Add 30 to 10.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{30±10}{8} when ± is minus. Subtract 10 from 30.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

The equation is now solved.

$$x=5 x=\frac{5}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-30x+4x^{2}=-50$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$4x^{2}-30x=-50$$

Divide both sides by 4.

$$\frac{4x^{2}-30x}{4}=-\frac{50}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{30}{4}\right)x=-\frac{50}{4}$$

Reduce the fraction \frac{-30}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{15}{2}x=-\frac{50}{4}$$

Reduce the fraction \frac{-50}{4} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{15}{2}, the coefficient of the x term, by 2 to get -\frac{15}{4}. Then add the square of -\frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{2}x+\left(-\frac{15}{4}\right)^{2}=-\frac{25}{2}+\left(-\frac{15}{4}\right)^{2}$$

Square -\frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=-\frac{25}{2}+\frac{225}{16}$$

Add -\frac{25}{2} to \frac{225}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{15}{2}x+\frac{225}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{15}{4}=\frac{5}{4} x-\frac{15}{4}=-\frac{5}{4}$$

Add \frac{15}{4} to both sides of the equation.

$$x=5 x=\frac{5}{2}$$