Question

use the quadratic formula to slove the equation -4x^2-3x+2=0

Answer 1

\[-4x^2-3x+2 =0 \rightarrow 4x^2+3x-2 =0 \rightarrow ax^2+bx+c =0\]
a= 4, b= 3 & c= -2
Now quadratic formula is given as;
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} =\frac{-3 \pm \sqrt{3^2-4\times 4 (-2)}}{2 \times 4} \]
\[=\frac{-3 \pm \sqrt{9+32}}{8} =\frac{-3 \pm \sqrt{41}}{8} \]
i.e. either \[x=\frac{-3 + \sqrt{41}}{8} --- or --x=\frac{-3- \sqrt{41}}{8}\]
are the required solutions of the given quadratic equation.

@NOOB_ZERO456

Answer 2

this is a multiple choice question
\[-\frac{ 3 }{ 8 }\pm \frac{ \sqrt{41} }{ 8 }\]
\[-\frac{ 8 }{ 3}\pm \frac{ \sqrt{82} }{ 8 }\]
\[-\frac{ 3 }{ 8 }\pm \frac{ \sqrt{20} }{2}\]
\[-\frac{ 3 }{ 4 }\pm \frac{ \sqrt{41} }{ 4 }\]
i don't understand this and would appreciate some help explaining this to me

Answer 3

First option is the correct answer
@NOOB_ZERO456

Answer 4

thanks you

Answer 5

\[ \huge x=\frac{−3± \sqrt{41}}{8} =\frac{−3}{8}± \frac{ \sqrt{41}}{8}\]
@NOOB_ZERO456 medal please