-4y^2+3y+7=0

Solution for -4y^2+3y+7=0 equation:

-4y^2+3y+7=0

a = -4; b = 3; c = +7;
Δ = b2-4ac
Δ = 32-4·(-4)·7
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*-4}=\frac{-14}{-8}
=1+3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*-4}=\frac{8}{-8}
=-1
$