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3x^2+46x+15=0
a = 3; b = 46; c = +15;
Δ = b2-4ac
Δ = 462-4·3·15
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-44}{2*3}=\frac{-90}{6} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+44}{2*3}=\frac{-2}{6} =-1/3 $
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