(3x+5)(2x-1)=9x^2-25

Solution for (3x+5)(2x-1)=9x^2-25 equation:

(3x+5)(2x-1)=9x^2-25

We move all terms to the left:

(3x+5)(2x-1)-(9x^2-25)=0

We get rid of parentheses

-9x^2+(3x+5)(2x-1)+25=0

We multiply parentheses ..

-9x^2+(+6x^2-3x+10x-5)+25=0

We get rid of parentheses

-9x^2+6x^2-3x+10x-5+25=0

We add all the numbers together, and all the variables

-3x^2+7x+20=0

a = -3; b = 7; c = +20;
Δ = b2-4ac
Δ = 72-4·(-3)·20
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*-3}=\frac{-24}{-6}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*-3}=\frac{10}{-6}
=-1+2/3
$