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436x^2+4=436
We move all terms to the left:
436x^2+4-(436)=0
We add all the numbers together, and all the variables
436x^2-432=0
a = 436; b = 0; c = -432;
Δ = b2-4ac
Δ = 02-4·436·(-432)
Δ = 753408
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{753408}=\sqrt{2304*327}=\sqrt{2304}*\sqrt{327}=48\sqrt{327}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{327}}{2*436}=\frac{0-48\sqrt{327}}{872} =-\frac{48\sqrt{327}}{872} =-\frac{6\sqrt{327}}{109} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{327}}{2*436}=\frac{0+48\sqrt{327}}{872} =\frac{48\sqrt{327}}{872} =\frac{6\sqrt{327}}{109} $
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