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4y(y+12)=26
We move all terms to the left:
4y(y+12)-(26)=0
We multiply parentheses
4y^2+48y-26=0
a = 4; b = 48; c = -26;
Δ = b2-4ac
Δ = 482-4·4·(-26)
Δ = 2720
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2720}=\sqrt{16*170}=\sqrt{16}*\sqrt{170}=4\sqrt{170}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{170}}{2*4}=\frac{-48-4\sqrt{170}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{170}}{2*4}=\frac{-48+4\sqrt{170}}{8} $
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