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16y^2+14y=69
We move all terms to the left:
16y^2+14y-(69)=0
a = 16; b = 14; c = -69;
Δ = b2-4ac
Δ = 142-4·16·(-69)
Δ = 4612
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4612}=\sqrt{4*1153}=\sqrt{4}*\sqrt{1153}=2\sqrt{1153}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{1153}}{2*16}=\frac{-14-2\sqrt{1153}}{32} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{1153}}{2*16}=\frac{-14+2\sqrt{1153}}{32} $
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