Solve 4x^2=27x+40

$$4 x ^ { 2 } = 27 x + 40$$

Solve for x

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$$x=8$$

Steps Using Factoring By Grouping

Subtract 27x from both sides.

$$4x^{2}-27x=40$$

Subtract 40 from both sides.

$$4x^{2}-27x-40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-27 ab=4\left(-40\right)=-160$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -160.

$$1,-160 2,-80 4,-40 5,-32 8,-20 10,-16$$

Calculate the sum for each pair.

$$1-160=-159 2-80=-78 4-40=-36 5-32=-27 8-20=-12 10-16=-6$$

The solution is the pair that gives sum -27.

$$a=-32 b=5$$

Rewrite 4x^{2}-27x-40 as \left(4x^{2}-32x\right)+\left(5x-40\right).

$$\left(4x^{2}-32x\right)+\left(5x-40\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(x-8\right)+5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(4x+5\right)$$

To find equation solutions, solve x-8=0 and 4x+5=0.

$$x=8 x=-\frac{5}{4}$$

Steps Using the Quadratic Formula

Subtract 27x from both sides.

$$4x^{2}-27x=40$$

Subtract 40 from both sides.

$$4x^{2}-27x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -27 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 4\left(-40\right)}}{2\times 4}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 4\left(-40\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-27\right)±\sqrt{729-16\left(-40\right)}}{2\times 4}$$

Multiply -16 times -40.

$$x=\frac{-\left(-27\right)±\sqrt{729+640}}{2\times 4}$$

Add 729 to 640.

$$x=\frac{-\left(-27\right)±\sqrt{1369}}{2\times 4}$$

Take the square root of 1369.

$$x=\frac{-\left(-27\right)±37}{2\times 4}$$

The opposite of -27 is 27.

$$x=\frac{27±37}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{27±37}{8}$$

Now solve the equation x=\frac{27±37}{8} when ± is plus. Add 27 to 37.

$$x=\frac{64}{8}$$

Divide 64 by 8.

$$x=8$$

Now solve the equation x=\frac{27±37}{8} when ± is minus. Subtract 37 from 27.

$$x=-\frac{10}{8}$$

Reduce the fraction \frac{-10}{8} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=8 x=-\frac{5}{4}$$

Steps for Completing the Square

Subtract 27x from both sides.

$$4x^{2}-27x=40$$

Divide both sides by 4.

$$\frac{4x^{2}-27x}{4}=\frac{40}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{27}{4}x=\frac{40}{4}$$

Divide 40 by 4.

$$x^{2}-\frac{27}{4}x=10$$

Divide -\frac{27}{4}, the coefficient of the x term, by 2 to get -\frac{27}{8}. Then add the square of -\frac{27}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{27}{4}x+\left(-\frac{27}{8}\right)^{2}=10+\left(-\frac{27}{8}\right)^{2}$$

Square -\frac{27}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{27}{4}x+\frac{729}{64}=10+\frac{729}{64}$$

Add 10 to \frac{729}{64}.

$$x^{2}-\frac{27}{4}x+\frac{729}{64}=\frac{1369}{64}$$

Factor x^{2}-\frac{27}{4}x+\frac{729}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{8}\right)^{2}=\frac{1369}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{8}\right)^{2}}=\sqrt{\frac{1369}{64}}$$

Simplify.

$$x-\frac{27}{8}=\frac{37}{8} x-\frac{27}{8}=-\frac{37}{8}$$

Add \frac{27}{8} to both sides of the equation.

$$x=8 x=-\frac{5}{4}$$