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Q. 105 E

Expert-verified
Found in: Page 885

### Intermediate Algebra

Book edition OER 2017
Author(s) OPENSTAX
Pages 1346 pages
ISBN 9780998625720

# In the following exercises, solve by completing the square.$2{x}^{2}+7x-15=0$

The solution is $x=-5,x=\frac{3}{2}$

See the step by step solution

## Step 1. Given information

Equation:$2{x}^{2}+7x-15=0$

## Step 2. Calculation

We have,

$2{x}^{2}+7x-15=0$

$2{x}^{2}+7x-15+15=0+15$

$\frac{2{x}^{2}+7x}{2}=\frac{15}{2}$

role="math" localid="1645178124973" ${x}^{2}+\frac{7x}{2}+{\left(\frac{1}{2}·\left(\frac{7}{2}\right)\right)}^{2}=\frac{15}{2}+{\left(\frac{1}{2}·\left(\frac{7}{2}\right)\right)}^{2}$

${x}^{2}+\frac{7x}{2}+{\left(\frac{7}{4}\right)}^{2}=\frac{15}{2}+{\left(\frac{7}{4}\right)}^{2}$

${x}^{2}+\frac{7x}{2}+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}$

${\left(x+\frac{7}{4}\right)}^{2}=\frac{169}{16}$

$\sqrt{{\left(x+\frac{7}{4}\right)}^{2}}=\sqrt{\frac{169}{16}}$

$x+\frac{7}{4}=±\frac{13}{4}$

$x+\frac{7}{4}-\frac{7}{4}=-\frac{13}{4}-\frac{7}{4},x+\frac{7}{4}-\frac{7}{4}=\frac{13}{4}-\frac{7}{4}$

$x=-\frac{20}{4},x=\frac{6}{4}$

$x=-5,x=\frac{3}{2}$