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12x^2+28x+15=0
a = 12; b = 28; c = +15;
Δ = b2-4ac
Δ = 282-4·12·15
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8}{2*12}=\frac{-36}{24} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8}{2*12}=\frac{-20}{24} =-5/6 $
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