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-8(2x+5)x=4
We move all terms to the left:
-8(2x+5)x-(4)=0
We multiply parentheses
-16x^2-40x-4=0
a = -16; b = -40; c = -4;
Δ = b2-4ac
Δ = -402-4·(-16)·(-4)
Δ = 1344
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{21}}{2*-16}=\frac{40-8\sqrt{21}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{21}}{2*-16}=\frac{40+8\sqrt{21}}{-32} $
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