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4x^2+10x-150=0
a = 4; b = 10; c = -150;
Δ = b2-4ac
Δ = 102-4·4·(-150)
Δ = 2500
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{2500}=50$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-50}{2*4}=\frac{-60}{8} =-7+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+50}{2*4}=\frac{40}{8} =5 $
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