Question

Help 1 question Step by step

Answer 1

Help anyone who can help me step by step

Answer 2

\(\bf 2x^2=-3x-5\implies 2x^2+3x+5=0\\ \quad \\
\textit{use the quadratic formula to get the values for "x"}\\ \quad \\
\text{quadratic formula}\\
x= \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\)

Answer 3

yea can you solve it for me steo by step @jdoe0001

Answer 4

\( \begin{array}{llll}
a&b&c\\
2x^2&+3x&+5=0
\end{array}\\ \quad \\
\bf
\text{quadratic formula}\\
x= \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\implies x= \cfrac{ - (3) \pm \sqrt { (3)^2 -4(2)(5)}}{2(2)}\\ \quad \\
x= \cfrac{ - 3 \pm \sqrt { 9 -40}}{4}\implies x= \cfrac{ - 3 \pm \sqrt { -31}}{4}
\)

Answer 5

okay thanks :)

Answer 6

yw

Answer 7

i have 5 more questions can you help me with the others?

Answer 8

i have answers just need to know if they are right?

Answer 9

I'll be dashing in a couple of mins

Answer 10

okay here they are
1.b
2.a
3.d
4.c
5.c

Answer 11

@jdoe0001 here they are above?

Answer 12

pretty straightforward, just use the quadratic formula on all, once you set them in the standard form, that is = 0, just like above

Answer 13

okay thanks