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3x^2+16x+21=9
We move all terms to the left:
3x^2+16x+21-(9)=0
We add all the numbers together, and all the variables
3x^2+16x+12=0
a = 3; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·3·12
Δ = 112
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{7}}{2*3}=\frac{-16-4\sqrt{7}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{7}}{2*3}=\frac{-16+4\sqrt{7}}{6} $
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