3x(3x-5)+(4x+3)=62

Solution for 3x(3x-5)+(4x+3)=62 equation:

3x(3x-5)+(4x+3)=62

We move all terms to the left:

3x(3x-5)+(4x+3)-(62)=0

We multiply parentheses

9x^2-15x+(4x+3)-62=0

We get rid of parentheses

9x^2-15x+4x+3-62=0

We add all the numbers together, and all the variables

9x^2-11x-59=0

a = 9; b = -11; c = -59;
Δ = b2-4ac
Δ = -112-4·9·(-59)
Δ = 2245
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{-(-11)-\sqrt{2245}}{2*9}=\frac{11-\sqrt{2245}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{2245}}{2*9}=\frac{11+\sqrt{2245}}{18}
$