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4x^2-35=0
a = 4; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·4·(-35)
Δ = 560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{35}}{2*4}=\frac{0-4\sqrt{35}}{8} =-\frac{4\sqrt{35}}{8} =-\frac{\sqrt{35}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{35}}{2*4}=\frac{0+4\sqrt{35}}{8} =\frac{4\sqrt{35}}{8} =\frac{\sqrt{35}}{2} $
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