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2y^2+14y-30=0
a = 2; b = 14; c = -30;
Δ = b2-4ac
Δ = 142-4·2·(-30)
Δ = 436
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{109}}{2*2}=\frac{-14-2\sqrt{109}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{109}}{2*2}=\frac{-14+2\sqrt{109}}{4} $
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