4y^2=-3y+10

Solution for 4y^2=-3y+10 equation:

4y^2=-3y+10

We move all terms to the left:

4y^2-(-3y+10)=0

We get rid of parentheses

4y^2+3y-10=0

a = 4; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·4·(-10)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*4}=\frac{-16}{8}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*4}=\frac{10}{8}
=1+1/4
$