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15=40x-16x^2
We move all terms to the left:
15-(40x-16x^2)=0
We get rid of parentheses
16x^2-40x+15=0
a = 16; b = -40; c = +15;
Δ = b2-4ac
Δ = -402-4·16·15
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{10}}{2*16}=\frac{40-8\sqrt{10}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{10}}{2*16}=\frac{40+8\sqrt{10}}{32} $
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