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=2250-35I-4I^2
We move all terms to the left:
-(2250-35I-4I^2)=0
We get rid of parentheses
4I^2+35I-2250=0
a = 4; b = 35; c = -2250;
Δ = b2-4ac
Δ = 352-4·4·(-2250)
Δ = 37225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{37225}=\sqrt{25*1489}=\sqrt{25}*\sqrt{1489}=5\sqrt{1489}$$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5\sqrt{1489}}{2*4}=\frac{-35-5\sqrt{1489}}{8} $$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5\sqrt{1489}}{2*4}=\frac{-35+5\sqrt{1489}}{8} $
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