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16x^2-40x=-5
We move all terms to the left:
16x^2-40x-(-5)=0
We add all the numbers together, and all the variables
16x^2-40x+5=0
a = 16; b = -40; c = +5;
Δ = b2-4ac
Δ = -402-4·16·5
Δ = 1280
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{5}}{2*16}=\frac{40-16\sqrt{5}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{5}}{2*16}=\frac{40+16\sqrt{5}}{32} $
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