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4x^2-1440=0
a = 4; b = 0; c = -1440;
Δ = b2-4ac
Δ = 02-4·4·(-1440)
Δ = 23040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{23040}=\sqrt{2304*10}=\sqrt{2304}*\sqrt{10}=48\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{10}}{2*4}=\frac{0-48\sqrt{10}}{8} =-\frac{48\sqrt{10}}{8} =-6\sqrt{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{10}}{2*4}=\frac{0+48\sqrt{10}}{8} =\frac{48\sqrt{10}}{8} =6\sqrt{10} $
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