# How do you solve 15 + 26d = -8d²?

Mar 31, 2018

#### Answer:

${d}_{1} = - \frac{3}{4} \mathmr{and} {d}_{2} = - \frac{5}{2}$

#### Explanation:

First, bring everything to one side.

$15 + 26 d = - 8 {d}^{2} | + 8 {d}^{2}$
$8 {d}^{2} + 26 d + 15 = 0$

Divide by the coefficient of ${d}^{2}$ which is $8$

8d^2+26d+15=0|:8
${d}^{2} + \frac{13}{4} d + \frac{15}{8} = 0$

Complete the square (${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$)

${d}^{2} + \frac{13}{4} \mathrm{dc} o l \mathmr{and} \left(b l u e\right) \left(+ {\left(\frac{13}{8}\right)}^{2} - {\left(\frac{13}{8}\right)}^{2}\right) + \frac{15}{8} = 0$
${\left(d + \frac{13}{8}\right)}^{2} - {\left(\frac{13}{8}\right)}^{2} + \frac{15}{8} = {\left(d + \frac{13}{8}\right)}^{2} - \frac{49}{64} = 0$

Solve for d

$0 = {\left(d + \frac{13}{8}\right)}^{2} - \frac{49}{64} | + \frac{49}{64} | \sqrt{} | - \frac{13}{8}$
$\pm \sqrt{\frac{49}{64}} - \frac{13}{8} = d$
$\pm \frac{7}{8} - \frac{13}{8} = d$
${d}_{1} = - \frac{3}{4} \mathmr{and} {d}_{2} = - \frac{5}{2}$