Answer:
The solution to the quadratic equations are:
[tex]x=-\frac{3+\sqrt{41}}{8},\:x=\frac{\sqrt{41}-3}{8}[/tex]
Step-by-step explanation:
Consider the provided quadratic equation.
[tex]-4x^2-3x+2=0[/tex]
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Substitute a=-4, b=-3 and c=2 in above formula.
[tex]x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\left(-4\right)2}}{2\left(-4\right)}[/tex]
[tex]x_{1,2}=\frac{3\pm\sqrt{9+32}}{-8}[/tex]
[tex]x_{1,2}=-\frac{3\pm\sqrt{41}}{8}[/tex]
Hence, the solution to the quadratic equations are:
[tex]x=-\frac{3+\sqrt{41}}{8},\:x=\frac{\sqrt{41}-3}{8}[/tex]
