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12x=-9x^2+10
We move all terms to the left:
12x-(-9x^2+10)=0
We get rid of parentheses
9x^2+12x-10=0
a = 9; b = 12; c = -10;
Δ = b2-4ac
Δ = 122-4·9·(-10)
Δ = 504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{14}}{2*9}=\frac{-12-6\sqrt{14}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{14}}{2*9}=\frac{-12+6\sqrt{14}}{18} $
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