Solve 30+2x=4x^2

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

Solve for x

$$x=3$$

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

Steps Using Factoring By Grouping

Subtract 4x^{2} from both sides.

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

Divide both sides by 2.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=1 ab=-2\times 15=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=6 b=-5$$

Rewrite -2x^{2}+x+15 as \left(-2x^{2}+6x\right)+\left(-5x+15\right).

$$\left(-2x^{2}+6x\right)+\left(-5x+15\right)$$

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

$$2x\left(-x+3\right)+5\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(2x+5\right)$$

To find equation solutions, solve -x+3=0 and 2x+5=0.

$$x=3 x=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -4.

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

Multiply 16 times 30.

$$x=\frac{-2±\sqrt{4+480}}{2\left(-4\right)}$$

Add 4 to 480.

$$x=\frac{-2±\sqrt{484}}{2\left(-4\right)}$$

Take the square root of 484.

$$x=\frac{-2±22}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2±22}{-8}$$

Now solve the equation x=\frac{-2±22}{-8} when ± is plus. Add -2 to 22.

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

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

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

Now solve the equation x=\frac{-2±22}{-8} when ± is minus. Subtract 22 from -2.

$$x=-\frac{24}{-8}$$

Divide -24 by -8.

$$x=3$$

The equation is now solved.

$$x=-\frac{5}{2} x=3$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

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

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

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

$$x^{2}-\frac{1}{2}x=-\frac{30}{-4}$$

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

$$x^{2}-\frac{1}{2}x=\frac{15}{2}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{15}{2}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{15}{2}+\frac{1}{16}$$

Add \frac{15}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{11}{4} x-\frac{1}{4}=-\frac{11}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=3 x=-\frac{5}{2}$$