Solve 5x^2=38x+16

$$5 x ^ { 2 } = 38 x + 16$$

Solve for x

$$x=-\frac{2}{5}=-0.4$$

$$x=8$$

Steps Using Factoring By Grouping

Subtract 38x from both sides.

$$5x^{2}-38x=16$$

Subtract 16 from both sides.

$$5x^{2}-38x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-38 ab=5\left(-16\right)=-80$$

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 -80.

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

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -38.

$$a=-40 b=2$$

Rewrite 5x^{2}-38x-16 as \left(5x^{2}-40x\right)+\left(2x-16\right).

$$\left(5x^{2}-40x\right)+\left(2x-16\right)$$

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

$$5x\left(x-8\right)+2\left(x-8\right)$$

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

$$\left(x-8\right)\left(5x+2\right)$$

To find equation solutions, solve x-8=0 and 5x+2=0.

$$x=8 x=-\frac{2}{5}$$

Steps Using the Quadratic Formula

Subtract 38x from both sides.

$$5x^{2}-38x=16$$

Subtract 16 from both sides.

$$5x^{2}-38x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -38 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-38\right)±\sqrt{\left(-38\right)^{2}-4\times 5\left(-16\right)}}{2\times 5}$$

Square -38.

$$x=\frac{-\left(-38\right)±\sqrt{1444-4\times 5\left(-16\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-38\right)±\sqrt{1444-20\left(-16\right)}}{2\times 5}$$

Multiply -20 times -16.

$$x=\frac{-\left(-38\right)±\sqrt{1444+320}}{2\times 5}$$

Add 1444 to 320.

$$x=\frac{-\left(-38\right)±\sqrt{1764}}{2\times 5}$$

Take the square root of 1764.

$$x=\frac{-\left(-38\right)±42}{2\times 5}$$

The opposite of -38 is 38.

$$x=\frac{38±42}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{38±42}{10}$$

Now solve the equation x=\frac{38±42}{10} when ± is plus. Add 38 to 42.

$$x=\frac{80}{10}$$

Divide 80 by 10.

$$x=8$$

Now solve the equation x=\frac{38±42}{10} when ± is minus. Subtract 42 from 38.

$$x=-\frac{4}{10}$$

Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=8 x=-\frac{2}{5}$$

Steps for Completing the Square

Subtract 38x from both sides.

$$5x^{2}-38x=16$$

Divide both sides by 5.

$$\frac{5x^{2}-38x}{5}=\frac{16}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{38}{5}x=\frac{16}{5}$$

Divide -\frac{38}{5}, the coefficient of the x term, by 2 to get -\frac{19}{5}. Then add the square of -\frac{19}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{38}{5}x+\left(-\frac{19}{5}\right)^{2}=\frac{16}{5}+\left(-\frac{19}{5}\right)^{2}$$

Square -\frac{19}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{38}{5}x+\frac{361}{25}=\frac{16}{5}+\frac{361}{25}$$

Add \frac{16}{5} to \frac{361}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{38}{5}x+\frac{361}{25}=\frac{441}{25}$$

Factor x^{2}-\frac{38}{5}x+\frac{361}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{19}{5}\right)^{2}=\frac{441}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{5}\right)^{2}}=\sqrt{\frac{441}{25}}$$

Simplify.

$$x-\frac{19}{5}=\frac{21}{5} x-\frac{19}{5}=-\frac{21}{5}$$

Add \frac{19}{5} to both sides of the equation.

$$x=8 x=-\frac{2}{5}$$