Question

Resolver a equação do 2-grau em irB) x2 _8x+12=0

Answer 1

Resposta:

x²-8x+12=0

$(-b±\sqrt{b²-4.a.c})/2.a$

$(-(-8)±\sqrt{(-8)²-4.1.12})/2.1$

$(8±\sqrt{64-60})/2$

$(8±\sqrt{4})/2$

$(8±2)/2$

x'=(8+2)/2=10/2=5

x"=(8-2)/2=6/2=3