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6x^2+16x+10=0
a = 6; b = 16; c = +10;
Δ = b2-4ac
Δ = 162-4·6·10
Δ = 16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*6}=\frac{-20}{12} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*6}=\frac{-12}{12} =-1 $
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