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3y^2+40y-675=0
a = 3; b = 40; c = -675;
Δ = b2-4ac
Δ = 402-4·3·(-675)
Δ = 9700
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9700}=\sqrt{100*97}=\sqrt{100}*\sqrt{97}=10\sqrt{97}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{97}}{2*3}=\frac{-40-10\sqrt{97}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{97}}{2*3}=\frac{-40+10\sqrt{97}}{6} $
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