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4a^2+28a-32=0
a = 4; b = 28; c = -32;
Δ = b2-4ac
Δ = 282-4·4·(-32)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-36}{2*4}=\frac{-64}{8} =-8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+36}{2*4}=\frac{8}{8} =1 $
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