How do you solve $p^{2} + 7 p + 8 = 0$ $p^2+7p+8=0$ using the quadratic formula?

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Answer 1 · 2017-07-19T03:25:44

See a solution process below:

For $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

Substituting $1$ $1$ for $a$ $a$ ; $7$ $7$ for $b$ $b$ and $8$ $8$ for $c$ $c$ gives:

$x = \frac{- 7 \pm \sqrt{7^{2} - (4 \cdot 1 \cdot 8)}}{2 \cdot 1}$ $x = (-7 +- sqrt(7^2 - (4 * 1 * 8)))/(2 * 1)$

$x = \frac{- 7 \pm \sqrt{49 - 32}}{2}$ $x = (-7 +- sqrt(49 - 32))/2$

$x = \frac{- 7 \pm \sqrt{17}}{2}$ $x = (-7 +- sqrt(17))/2$

Or

$x = \frac{- 7 + \sqrt{17}}{2}$ $x = (-7 + sqrt(17))/2$ and $x = \frac{- 7 - \sqrt{17}}{2}$ $x = (-7 - sqrt(17))/2$

How do you solve p2+7p+8=0p^2+7p+8=0 using the quadratic formula?