(3x+4)(4x+5)=Y

Solution for (3x+4)(4x+5)=Y equation:

(3x+4)(4x+5)=

We move all terms to the left:

(3x+4)(4x+5)-()=0

We add all the numbers together, and all the variables

(3x+4)(4x+5)=0

We multiply parentheses ..

(+12x^2+15x+16x+20)=0

We get rid of parentheses

12x^2+15x+16x+20=0

We add all the numbers together, and all the variables

12x^2+31x+20=0

a = 12; b = 31; c = +20;
Δ = b2-4ac
Δ = 312-4·12·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-1}{2*12}=\frac{-32}{24}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+1}{2*12}=\frac{-30}{24}
=-1+1/4
$