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10x+2x^2=61
We move all terms to the left:
10x+2x^2-(61)=0
a = 2; b = 10; c = -61;
Δ = b2-4ac
Δ = 102-4·2·(-61)
Δ = 588
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14\sqrt{3}}{2*2}=\frac{-10-14\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14\sqrt{3}}{2*2}=\frac{-10+14\sqrt{3}}{4} $
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