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23x+5=-12x^2
We move all terms to the left:
23x+5-(-12x^2)=0
We get rid of parentheses
12x^2+23x+5=0
a = 12; b = 23; c = +5;
Δ = b2-4ac
Δ = 232-4·12·5
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*12}=\frac{-40}{24} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*12}=\frac{-6}{24} =-1/4 $
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