Question

x²+3x-4=0 e em forma bhaskara

Answer 1

$x^{2} +3x-4=0$

a=1     b= 3     c = -4     Δ=b²-4ac = 3²-4*1*(-4) = 9+16     Δ=25     √Δ = √25 = 5

$x= \frac{-b+- \sqrt{Delta} }{2a} \\ \\ x= \frac{-3+-5}{2} \\ \\ x'= \frac{-3-5}{2}= \frac{-8}{2} = -4 \\ \\ x'= \frac{-3+5}{2}= \frac{2}{2} = 1$

S={-4, 1}

Answer 2

a=1
b=3
c=-4

Δ=b²-4ac
Δ=3²-4*1*(-4)
Δ=9+16
Δ=25

X'=(-b+√Δ)/2a
X'=(-3+$\sqrt{25}$)2a
X'=(-3+5)/2*1
X'=-3-5/2
X'=-8/2
X'=-4

X"=-3+5/2
X"=2/2
X"=1

Tirando a prova:
Para X'=-4

x²+3x-4=0
-4²+3*(-4)-4=0
16+-12-4=0
16-16=0
0=0

Para X"=1
1²+3*1-4=0
1+3-4=0
0=0