Solve 5x^2=32x+21

$$5 x ^ { 2 } = 32 x + 21$$

Solve for x

$$x=-\frac{3}{5}=-0.6$$

$$x=7$$

Steps Using Factoring By Grouping

Subtract 32x from both sides.

$$5x^{2}-32x=21$$

Subtract 21 from both sides.

$$5x^{2}-32x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=-32 ab=5\left(-21\right)=-105$$

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 -105.

$$1,-105 3,-35 5,-21 7,-15$$

Calculate the sum for each pair.

$$1-105=-104 3-35=-32 5-21=-16 7-15=-8$$

The solution is the pair that gives sum -32.

$$a=-35 b=3$$

Rewrite 5x^{2}-32x-21 as \left(5x^{2}-35x\right)+\left(3x-21\right).

$$\left(5x^{2}-35x\right)+\left(3x-21\right)$$

Factor out 5x in the first and 3 in the second group.

$$5x\left(x-7\right)+3\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(5x+3\right)$$

To find equation solutions, solve x-7=0 and 5x+3=0.

$$x=7 x=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Subtract 32x from both sides.

$$5x^{2}-32x=21$$

Subtract 21 from both sides.

$$5x^{2}-32x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -32 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 5\left(-21\right)}}{2\times 5}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 5\left(-21\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-32\right)±\sqrt{1024-20\left(-21\right)}}{2\times 5}$$

Multiply -20 times -21.

$$x=\frac{-\left(-32\right)±\sqrt{1024+420}}{2\times 5}$$

Add 1024 to 420.

$$x=\frac{-\left(-32\right)±\sqrt{1444}}{2\times 5}$$

Take the square root of 1444.

$$x=\frac{-\left(-32\right)±38}{2\times 5}$$

The opposite of -32 is 32.

$$x=\frac{32±38}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{32±38}{10}$$

Now solve the equation x=\frac{32±38}{10} when ± is plus. Add 32 to 38.

$$x=\frac{70}{10}$$

Divide 70 by 10.

$$x=7$$

Now solve the equation x=\frac{32±38}{10} when ± is minus. Subtract 38 from 32.

$$x=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=7 x=-\frac{3}{5}$$

Steps for Completing the Square

Subtract 32x from both sides.

$$5x^{2}-32x=21$$

Divide both sides by 5.

$$\frac{5x^{2}-32x}{5}=\frac{21}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{32}{5}x=\frac{21}{5}$$

Divide -\frac{32}{5}, the coefficient of the x term, by 2 to get -\frac{16}{5}. Then add the square of -\frac{16}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{5}x+\left(-\frac{16}{5}\right)^{2}=\frac{21}{5}+\left(-\frac{16}{5}\right)^{2}$$

Square -\frac{16}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{5}x+\frac{256}{25}=\frac{21}{5}+\frac{256}{25}$$

Add \frac{21}{5} to \frac{256}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{5}x+\frac{256}{25}=\frac{361}{25}$$

Factor x^{2}-\frac{32}{5}x+\frac{256}{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}.

$$\left(x-\frac{16}{5}\right)^{2}=\frac{361}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{5}\right)^{2}}=\sqrt{\frac{361}{25}}$$

Simplify.

$$x-\frac{16}{5}=\frac{19}{5} x-\frac{16}{5}=-\frac{19}{5}$$

Add \frac{16}{5} to both sides of the equation.

$$x=7 x=-\frac{3}{5}$$