\(4 x^{2} +20x+25=0\)
\(a x^{2} +bx+c=0\)
Δ=\(b^{2} -4ac\)
Δ=\(20^{2} - 4*4*25\)
Δ=400 - 400 = 0
if
Δ > 0 equation has 2 solution
Δ = 0 equation has 1 solution
Δ < 0 no real solution
\(x= \frac{-b}{2a}\)
\(x_{0} = \frac{-20}{2*4} = \frac{-20}{8} = -2.5\)
