Question

solve 2x^2+3x=-5 solve 2x^2+3x=-5 @Mathematics

Answer 1

2x^2 + 3x + 5 = 0
a = 2 , b = 3 , c = 5
Sub it into the quadratic formula

Answer 2

\[x^2+\frac{3}{2}x=\frac{-5}{2}\]
\[x^2+\frac{3}{2}x+(\frac{3}{2 \cdot 2})^2=\frac{-5}{2}+(\frac{3}{2 \cdot 2})^2\]
\[x^2+\frac{3}{2}x+(\frac{3}{4})^2=\frac{-5}{2}+\frac{9}{16}\]
\[(x+\frac{3}{4})^2=\frac{-40+9}{16}\]

Answer 3

\[x+\frac{3}{4}= \pm \sqrt{\frac{-31}{16}}\]

Answer 4

\[x=-\frac{3}{4} \pm \frac{\sqrt{-31}}{4}=\frac{-3 \pm i \sqrt{31}}{4}\]

Answer 5

2x^2+3x=-5
x^2+3x/2 = -5/2
add 9/16 to both sides so we can factor
x^2+3x/2+9/16=-31/16
(x+3/4)^2=-31/16
\[|x+\frac{3}{4}|=\sqrt{-\frac{31}{16}}\]
\[|x+\frac{3}{4}|=\frac{i \sqrt{31}}{4}\]
\[x+\frac{3}{4}=\pm\frac{i \sqrt{31}}{4}\]
\[x=-\frac{1}{4}i(\sqrt{31}-3i)\]
and
\[x=\frac{1}{4}i(\sqrt{31}+3i)\]