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3d^2+16d+13=0
a = 3; b = 16; c = +13;
Δ = b2-4ac
Δ = 162-4·3·13
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-10}{2*3}=\frac{-26}{6} =-4+1/3 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+10}{2*3}=\frac{-6}{6} =-1 $
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