8x^2-232x+1584=0

Solution for 8x^2-232x+1584=0 equation:

8x^2-232x+1584=0

a = 8; b = -232; c = +1584;
Δ = b2-4ac
Δ = -2322-4·8·1584
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-232)-56}{2*8}=\frac{176}{16}
=11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-232)+56}{2*8}=\frac{288}{16}
=18
$