49y^2+42y-16=0

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Solution for 49y^2+42y-16=0 equation:



49y^2+42y-16=0
a = 49; b = 42; c = -16;
Δ = b2-4ac
Δ = 422-4·49·(-16)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-70}{2*49}=\frac{-112}{98} =-1+1/7 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+70}{2*49}=\frac{28}{98} =2/7 $

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