-(2x+1)(2x+5)=3x-14

Solution for -(2x+1)(2x+5)=3x-14 equation:

-(2x+1)(2x+5)=3x-14

We move all terms to the left:

-(2x+1)(2x+5)-(3x-14)=0

We get rid of parentheses

-(2x+1)(2x+5)-3x+14=0

We multiply parentheses ..

-(+4x^2+10x+2x+5)-3x+14=0

We get rid of parentheses

-4x^2-10x-2x-3x-5+14=0

We add all the numbers together, and all the variables

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

a = -4; b = -15; c = +9;
Δ = b2-4ac
Δ = -152-4·(-4)·9
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*-4}=\frac{15-3\sqrt{41}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*-4}=\frac{15+3\sqrt{41}}{-8}
$