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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2x^2+x-10x-5)-4=0

We get rid of parentheses

2x^2+x-10x-5-4=0

We add all the numbers together, and all the variables

2x^2-9x-9=0

a = 2; b = -9; c = -9;
Δ = b2-4ac
Δ = -92-4·2·(-9)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{17}}{2*2}=\frac{9-3\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{17}}{2*2}=\frac{9+3\sqrt{17}}{4}
$