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4y^2+48y-10=0
a = 4; b = 48; c = -10;
Δ = b2-4ac
Δ = 482-4·4·(-10)
Δ = 2464
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2464}=\sqrt{16*154}=\sqrt{16}*\sqrt{154}=4\sqrt{154}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{154}}{2*4}=\frac{-48-4\sqrt{154}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{154}}{2*4}=\frac{-48+4\sqrt{154}}{8} $
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