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10x^2+10=29x
We move all terms to the left:
10x^2+10-(29x)=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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-21}{2*10}=\frac{8}{20} =2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+21}{2*10}=\frac{50}{20} =2+1/2 $
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