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10x^2+29x+10=0
a = 10; b = 29; c = +10;
Δ = b2-4ac
Δ = 292-4·10·10
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*10}=\frac{-50}{20} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*10}=\frac{-8}{20} =-2/5 $
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