Solve 4x^2=40-6x

$$4 x ^ { 2 } = 40 - 6 x$$

Solve for x

$$x=-4$$

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

Steps Using Factoring By Grouping

Subtract 40 from both sides.

$$4x^{2}-40=-6x$$

Add 6x to both sides.

$$4x^{2}-40+6x=0$$

Divide both sides by 2.

$$2x^{2}-20+3x=0$$

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

$$2x^{2}+3x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-20\right)=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=-5 b=8$$

Rewrite 2x^{2}+3x-20 as \left(2x^{2}-5x\right)+\left(8x-20\right).

$$\left(2x^{2}-5x\right)+\left(8x-20\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(2x-5\right)+4\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(x+4\right)$$

To find equation solutions, solve 2x-5=0 and x+4=0.

$$x=\frac{5}{2} x=-4$$

Steps Using the Quadratic Formula

Subtract 40 from both sides.

$$4x^{2}-40=-6x$$

Add 6x to both sides.

$$4x^{2}-40+6x=0$$

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.

$$4x^{2}+6x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 6 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-40\right)}}{2\times 4}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 4\left(-40\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16\left(-40\right)}}{2\times 4}$$

Multiply -16 times -40.

$$x=\frac{-6±\sqrt{36+640}}{2\times 4}$$

Add 36 to 640.

$$x=\frac{-6±\sqrt{676}}{2\times 4}$$

Take the square root of 676.

$$x=\frac{-6±26}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-6±26}{8}$$

Now solve the equation x=\frac{-6±26}{8} when ± is plus. Add -6 to 26.

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

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{-6±26}{8} when ± is minus. Subtract 26 from -6.

$$x=-\frac{32}{8}$$

Divide -32 by 8.

$$x=-4$$

The equation is now solved.

$$x=\frac{5}{2} x=-4$$

Steps for Completing the Square

Add 6x to both sides.

$$4x^{2}+6x=40$$

Divide both sides by 4.

$$\frac{4x^{2}+6x}{4}=\frac{40}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{6}{4}x=\frac{40}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{2}x=\frac{40}{4}$$

Divide 40 by 4.

$$x^{2}+\frac{3}{2}x=10$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=10+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=10+\frac{9}{16}$$

Add 10 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{169}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{13}{4} x+\frac{3}{4}=-\frac{13}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{5}{2} x=-4$$