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4x^2+100x+624=0
a = 4; b = 100; c = +624;
Δ = b2-4ac
Δ = 1002-4·4·624
Δ = 16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-4}{2*4}=\frac{-104}{8} =-13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+4}{2*4}=\frac{-96}{8} =-12 $
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