Question

resolva em ir (x-3) (2x+5)=0​

Answer 1

Explicação passo-a-passo:

(x-3)(2x+5)=0

2x²+5x-6x-15=0

2x²-x-15=0

a=2

b=-1

c=-15

∆=b²-4ac

∆=(-1)²-4*2*-15

∆=1+120

∆=121

-b±√∆/2a

1±√121/2*2

1±11/4

x¹=1+11/4=12/4=>3

x²=1-11/4=-10/4(÷2)=>-5/2

Answer 2

Resposta:

S={-5/2, 3}

Explicação passo-a-passo:

$(x - 3).(2x + 5) = 0$

$2x { }^{2} + 5x - 6x - 15 = 0 \\ 2x {}^{2} - x - 15 = 0$

$a = 2 \: \: \: b = - 1 \: \: \: c = - 15$

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

$deslta = ( - 1) {}^{2} - 4.2. - 15 = 1 + 120 = 121$

$\sqrt{121} = 11$

$x1 = \frac{ - ( - 1) + 11}{2.2} = \frac{1 + 11}{4} = \frac{12}{4} = 3$

$x2 = \frac{ - ( - 1) - 11}{2.2} = \frac{1 - 11}{4} = \frac{ - 10}{4} = \frac{ - 5}{2}$