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4(-305+10x+2x^2)=0
We multiply parentheses
8x^2+40x-1220=0
a = 8; b = 40; c = -1220;
Δ = b2-4ac
Δ = 402-4·8·(-1220)
Δ = 40640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40640}=\sqrt{64*635}=\sqrt{64}*\sqrt{635}=8\sqrt{635}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{635}}{2*8}=\frac{-40-8\sqrt{635}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{635}}{2*8}=\frac{-40+8\sqrt{635}}{16} $
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