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10x^2+68x-45=0
a = 10; b = 68; c = -45;
Δ = b2-4ac
Δ = 682-4·10·(-45)
Δ = 6424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6424}=\sqrt{4*1606}=\sqrt{4}*\sqrt{1606}=2\sqrt{1606}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-2\sqrt{1606}}{2*10}=\frac{-68-2\sqrt{1606}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+2\sqrt{1606}}{2*10}=\frac{-68+2\sqrt{1606}}{20} $
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