Algebra Examples

$4 x^{2} + 25 = 20 x$

Step 1

Subtract $20 x$ from both sides of the equation.

$4 x^{2} + 25 - 20 x = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 4$ , $b = - 20$ , and $c = 25$ into the quadratic formula and solve for $x$ .

$\frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \cdot (4 \cdot 25)}}{2 \cdot 4}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 20$ to the power of $2$ .

$x = \frac{20 \pm \sqrt{400 - 4 \cdot 4 \cdot 25}}{2 \cdot 4}$

Step 4.1.2

Multiply $- 4 \cdot 4 \cdot 25$ .

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Step 4.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{20 \pm \sqrt{400 - 16 \cdot 25}}{2 \cdot 4}$

Step 4.1.2.2

Multiply $- 16$ by $25$ .

$x = \frac{20 \pm \sqrt{400 - 400}}{2 \cdot 4}$

Step 4.1.3

Subtract $400$ from $400$ .

$x = \frac{20 \pm \sqrt{0}}{2 \cdot 4}$

Step 4.1.4

Rewrite $0$ as $0^{2}$ .

$x = \frac{20 \pm \sqrt{0^{2}}}{2 \cdot 4}$

Step 4.1.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{20 \pm 0}{2 \cdot 4}$

Step 4.1.6

$20$ plus or minus $0$ is $20$ .

$x = \frac{20}{2 \cdot 4}$

Step 4.2

Multiply $2$ by $4$ .

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

Step 4.3

Cancel the common factor of $20$ and $8$ .

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Step 4.3.1

Factor $4$ out of $20$ .

$x = \frac{4 (5)}{8}$

Step 4.3.2

Cancel the common factors.

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Step 4.3.2.1

Factor $4$ out of $8$ .

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

Step 4.3.2.2

Cancel the common factor.

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

Step 4.3.2.3

Rewrite the expression.

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

Step 5

The final answer is the combination of both solutions.

$x = \frac{5}{2}$ Double roots