Question

m²+m-6=0 Ajutati-ma va rog sa il rezolv. Urgent!!!!!!!

Answer 1

ax²+bx+c=0, Δ=b²-4ac, x1,2=(-b+-√Δ)/2a
m²+m-6=0
Δ=1-4(-6)=1+24=25
m1=(-1+5)/2=2
m2=(-1-5)/2=-3

Answer 2

$\displaystyle m^2+m-6=0 \\ a=1,~b=1,~c=-6 \\ \Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-6)=1+24=25\ \textgreater \ 0 \\ m_1= \frac{-1- \sqrt{25} }{2 \cdot 1} = \frac{-1-5}{2} = \frac{-6}{2} =-3 \\ m_2= \frac{-1+ \sqrt{25} }{2 \cdot 1} = \frac{-1+5}{2} = \frac{4}{2} =2$