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4z^2+31-24z=0
a = 4; b = -24; c = +31;
Δ = b2-4ac
Δ = -242-4·4·31
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{5}}{2*4}=\frac{24-4\sqrt{5}}{8} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{5}}{2*4}=\frac{24+4\sqrt{5}}{8} $
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