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10x^2=3x+3
We move all terms to the left:
10x^2-(3x+3)=0
We get rid of parentheses
10x^2-3x-3=0
a = 10; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·10·(-3)
Δ = 129
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{129}}{2*10}=\frac{3-\sqrt{129}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*10}=\frac{3+\sqrt{129}}{20} $
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