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y^2-10=-3y
We move all terms to the left:
y^2-10-(-3y)=0
We get rid of parentheses
y^2+3y-10=0
a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*1}=\frac{-10}{2} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2} =2 $
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