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2d^2+16d-10=0
a = 2; b = 16; c = -10;
Δ = b2-4ac
Δ = 162-4·2·(-10)
Δ = 336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{21}}{2*2}=\frac{-16-4\sqrt{21}}{4} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{21}}{2*2}=\frac{-16+4\sqrt{21}}{4} $
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