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8x(2x+5)=10
We move all terms to the left:
8x(2x+5)-(10)=0
We multiply parentheses
16x^2+40x-10=0
a = 16; b = 40; c = -10;
Δ = b2-4ac
Δ = 402-4·16·(-10)
Δ = 2240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2240}=\sqrt{64*35}=\sqrt{64}*\sqrt{35}=8\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{35}}{2*16}=\frac{-40-8\sqrt{35}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{35}}{2*16}=\frac{-40+8\sqrt{35}}{32} $
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