(1/5)³ˣ²⁺ˣ⁻¹³=125
(5⁻¹)³ˣ²⁺ˣ⁻¹³=5³
(5)⁻³ˣ²⁻ˣ⁺¹³=5³⇒-3x²-x+13=3(baze egale->am egalat exponentii)
vom inmulti cu (-1)ambii membri⇒3x²+x-13=-3
3x²+x-10=0(a=3;b=1;c=-10)
Δ=b²-4ac=1²-4·3·(-10)=1+120=121⇒√Δ=11
x₁=(-b+√Δ)/2a=(-1+11)/2·3=10/6=5/3
x₂=(-b-√Δ)2a=(-1-11)/2·3=-12÷6=-2
