Question

equação de 2º grau:
a) x²+13x-48=0
b) x²+17x+60=0

Answer 1

a) x²+13x-48=0

∆ = 13² - 4.1.(-48)
∆ = 169 + 192
∆ = 361

x = (-13 +-√361)/2.1
x' = (-13 -19)/2 = - 32/2 = -16
x" = (-13 +19)/2 = 6/2 = 3

b) x²+17x+60=0

∆ = 17² - 4.1.60
∆ = 289 - 240
∆ = 49

x = (-17 +-√49)/2.1
x' = (-17 -7)/2 =-24/2 = -12
x" = (-17 +7)/2 = -10/2 = -5

Answer 2

${b}^{2} - 4 \times a \times c \\ \\ {13}^{2} - 4 \times 1 \times ( - 48) \\ {13}^{2} + 192 \\ 168 + 192 = 361 \\ delt = 361$
$- b - + \sqrt{361 \div 2 \times a} \\ x1 = - 13 + 19 \div 2 = 3 \\ x2 = - 13 - 19 \div 2 = 16$