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9x^2+4x=16
We move all terms to the left:
9x^2+4x-(16)=0
a = 9; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·9·(-16)
Δ = 592
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{37}}{2*9}=\frac{-4-4\sqrt{37}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{37}}{2*9}=\frac{-4+4\sqrt{37}}{18} $
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