5x2 - 23x - 10 = 0
A.x = -2/5,5
b. x = -10,5
C.x = -2, 25
Applying the formula of quadratic equation
[tex]x=\frac{-b\pm\sqrt[\square]{b^2-4ac}}{2a}[/tex]we have
a=5
b=-23
c=-10
substitute in the formula
[tex]\begin{gathered} x=\frac{-(-23)\pm\sqrt[\square]{(-23)^2-4(5)(-10)}}{2(5)} \\ \\ x=\frac{23\pm\sqrt[\square]{729}}{10} \\ \\ x=\frac{23\pm27}{10} \\ x=5 \\ x=-0.4 \end{gathered}[/tex]answer is option A
