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-2j^2+36j+10=0
a = -2; b = 36; c = +10;
Δ = b2-4ac
Δ = 362-4·(-2)·10
Δ = 1376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1376}=\sqrt{16*86}=\sqrt{16}*\sqrt{86}=4\sqrt{86}$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{86}}{2*-2}=\frac{-36-4\sqrt{86}}{-4} $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{86}}{2*-2}=\frac{-36+4\sqrt{86}}{-4} $
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