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y^2+10y-35=0
a = 1; b = 10; c = -35;
Δ = b2-4ac
Δ = 102-4·1·(-35)
Δ = 240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{15}}{2*1}=\frac{-10-4\sqrt{15}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{15}}{2*1}=\frac{-10+4\sqrt{15}}{2} $
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