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

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

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

We move all terms to the left:

-2y^2+7y+3-(3)=0

We add all the numbers together, and all the variables

-2y^2+7y=0

a = -2; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-2)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-2}=\frac{-14}{-4}
=3+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-2}=\frac{0}{-4}
=0
$