Question

x2-7×+12=0
me ajudemm​

Answer 1

$\large\boxed{\begin{array}{l}\sf x^2-7x+12=0\\\sf a=1\quad b=-7\quad c=12\\\sf\Delta=b^2-4\cdot a\cdot c\\\sf\Delta=(-7)^2-4\cdot1\cdot12\\\sf\Delta=49- 48\\\sf\Delta=1\\\sf x=\dfrac{-b\pm\sqrt\Delta}{2\cdot a}\\\\\sf x=\dfrac{-(-7)\pm\sqrt1}{2\cdot 1}\\\\\sf x=\dfrac{7\pm1}{2}\begin{cases}\sf x'=\dfrac{7+1}{2}=\dfrac{8}{2}=4\\\\\sf x''=\dfrac{7-1}{2}=\dfrac{6}{2}=3\end{cases}\\\sf S=\{3\,,\,4\}\end{array}}$

Answer 2

Explicação passo-a-passo:

$x {}^{2} - 7x + 12 = 0$

$a = 1 \: \: \: \: b = - 7 \: \: \: \: c = 12$

$delta = b {}^{2} - 4ac$

$delta = ( - 7) {}^{2} - 4.1.12 = 49 - 48 = 1$

$\sqrt{1} = 1$

$x1 = \frac{ - ( - 7) + 1}{2.1} = \frac{7 + 1}{.2} = \frac{8}{2} = 4$

$x2 = \frac{ - ( - - 7) - 1}{2.1} = \frac{7 - 1}{2} = \frac{6}{2} = 3$