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

Solution for 4(x^2+x-4)=0 equation:

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

We multiply parentheses

4x^2+4x-16=0

a = 4; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·4·(-16)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{17}}{2*4}=\frac{-4-4\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{17}}{2*4}=\frac{-4+4\sqrt{17}}{8}
$