Algebra Examples

$4 b^{2} + 8 b + 7 = 4$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Subtract $4$ from both sides of the equation.

$4 b^{2} + 8 b + 7 - 4 = 0$

Step 1.2

Subtract $4$ from $7$ .

$4 b^{2} + 8 b + 3 = 0$

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 $a = 4$ , $b = 8$ , and $c = 3$ into the quadratic formula and solve for $b$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (4 \cdot 3)}}{2 \cdot 4}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $8$ to the power of $2$ .

$b = \frac{- 8 \pm \sqrt{64 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Step 4.1.2

Multiply $- 4 \cdot 4 \cdot 3$ .

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Step 4.1.2.1

Multiply $- 4$ by $4$ .

$b = \frac{- 8 \pm \sqrt{64 - 16 \cdot 3}}{2 \cdot 4}$

Step 4.1.2.2

Multiply $- 16$ by $3$ .

$b = \frac{- 8 \pm \sqrt{64 - 48}}{2 \cdot 4}$

Step 4.1.3

Subtract $48$ from $64$ .

$b = \frac{- 8 \pm \sqrt{16}}{2 \cdot 4}$

Step 4.1.4

Rewrite $16$ as $4^{2}$ .

$b = \frac{- 8 \pm \sqrt{4^{2}}}{2 \cdot 4}$

Step 4.1.5

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

$b = \frac{- 8 \pm 4}{2 \cdot 4}$

Step 4.2

Multiply $2$ by $4$ .

$b = \frac{- 8 \pm 4}{8}$

Step 4.3

Simplify $\frac{- 8 \pm 4}{8}$ .

$b = \frac{- 2 \pm 1}{2}$

Step 5

The final answer is the combination of both solutions.

$b = - \frac{1}{2}, - \frac{3}{2}$