(5x+2)(3x-1)=4x^2+5

Solution for (5x+2)(3x-1)=4x^2+5 equation:

(5x+2)(3x-1)=4x^2+5

We move all terms to the left:

(5x+2)(3x-1)-(4x^2+5)=0

We get rid of parentheses

-4x^2+(5x+2)(3x-1)-5=0

We multiply parentheses ..

-4x^2+(+15x^2-5x+6x-2)-5=0

We get rid of parentheses

-4x^2+15x^2-5x+6x-2-5=0

We add all the numbers together, and all the variables

11x^2+x-7=0

a = 11; b = 1; c = -7;
Δ = b2-4ac
Δ = 12-4·11·(-7)
Δ = 309
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{309}}{2*11}=\frac{-1-\sqrt{309}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{309}}{2*11}=\frac{-1+\sqrt{309}}{22}
$