Question

Determine as raizes da equaçao abaixo

Answer 1

$-x^2+12x+20=0$

Utilizando a fórmula de Bhaskara, anexa
a=-1

b=12

c=20

$delta = 12^2-4(-1).20=144+80=224 \\ \\ \sqrt{delta} = \sqrt{224} = \sqrt{2^5.7} =2^2 \sqrt{2.7} =4 \sqrt{14} \\ \\ x_{1} = \frac{-12+4 \sqrt{14} }{2.(-1)} = \frac{-12}{-2} + \frac{4 \sqrt{14} }{-2} =6-2 \sqrt{14}=2(3- \sqrt{14} ) \\ \\ x_{2} = \frac{-12-4 \sqrt{14} }{2.(-1)} = \frac{-12}{-2} - \frac{4 \sqrt{14} }{-2} =6+2 \sqrt{14} =2(3+ \sqrt{14} )$

As ráizes são $x_{1} =6- 2\sqrt{14}$ e$x_{2} =6-2 \sqrt{14}$