2z^2=3z+5

Solution for 2z^2=3z+5 equation:

2z^2=3z+5

We move all terms to the left:

2z^2-(3z+5)=0

We get rid of parentheses

2z^2-3z-5=0

a = 2; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·2·(-5)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*2}=\frac{-4}{4}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*2}=\frac{10}{4}
=2+1/2
$