Question

quanto é -x²+13x-42=0?

Answer 1

-x²+13x-42=0

x²-13x+42=0

Δ=b²-4ac
Δ=(-13)²-4(42)
Δ=169-168
Δ=1
√Δ=√1

$x= \frac{-(-13)\pm \sqrt{1} }{2} = \frac{13\pm1}{2}$

$x'= \frac{13+1}{2} = \frac{14}{2} =7$

$x"= \frac{13-1}{2}= \frac{12}{2}=6$

S={6,7}