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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+8x^2-4x+6x-3)-5=0

We get rid of parentheses

8x^2-4x+6x-3-5=0

We add all the numbers together, and all the variables

8x^2+2x-8=0

a = 8; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·8·(-8)
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{65}}{2*8}=\frac{-2-2\sqrt{65}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{65}}{2*8}=\frac{-2+2\sqrt{65}}{16}
$