(3a^2+2a-5)/(a^2+3a-4)=5

Solution for (3a^2+2a-5)/(a^2+3a-4)=5 equation:

(3a^2+2a-5)/(a^2+3a-4)=5

We move all terms to the left:

(3a^2+2a-5)/(a^2+3a-4)-(5)=0


Domain of the equation: (a^2+3a-4)!=0

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

a^2+3a!=4

a∈R


We multiply all the terms by the denominator


(3a^2+2a-5)-5*(a^2+3a-4)=0

We multiply parentheses

-5a^2+(3a^2+2a-5)-15a+20=0

We get rid of parentheses

-5a^2+3a^2+2a-15a-5+20=0

We add all the numbers together, and all the variables

-2a^2-13a+15=0

a = -2; b = -13; c = +15;
Δ = b2-4ac
Δ = -132-4·(-2)·15
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*-2}=\frac{-4}{-4}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*-2}=\frac{30}{-4}
=-7+1/2
$