(2b^2+3b-5)/(b-3)=0

Solution for (2b^2+3b-5)/(b-3)=0 equation:

(2b^2+3b-5)/(b-3)=0


Domain of the equation: (b-3)!=0

We move all terms containing b to the left, all other terms to the right

b!=3

b∈R


We multiply all the terms by the denominator


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

We get rid of parentheses

2b^2+3b-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:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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