Respuesta :
7x^2+8x-12=2x^2+14x-4
⇒ 7x^2 - 2x^2 + 8x - 14x -12 +4 = 0
⇒ 5x^2 - 6x - 8 = 0
b^2 - 4ac = (-6)^2 - 4 * 5 * (-8) = 36 + 160 = 196 > 0
So there are two real roots.
Hope it helps!
⇒ 7x^2 - 2x^2 + 8x - 14x -12 +4 = 0
⇒ 5x^2 - 6x - 8 = 0
b^2 - 4ac = (-6)^2 - 4 * 5 * (-8) = 36 + 160 = 196 > 0
So there are two real roots.
Hope it helps!
I hope this helps you
7x^2+8x-12-2x^2-14x+4=0
5x^2-6x-8=0
a=5 b= -6 c= -8
disctirminant =b^2-4ac
disctirminant =(-6)^2-4.5. (-8)
disctirminant =36+160
disctirminant =196
x1=6+ square root of 196/2.5
x1=3+2 square root of 3/5
x2=3- square root of 3/5
7x^2+8x-12-2x^2-14x+4=0
5x^2-6x-8=0
a=5 b= -6 c= -8
disctirminant =b^2-4ac
disctirminant =(-6)^2-4.5. (-8)
disctirminant =36+160
disctirminant =196
x1=6+ square root of 196/2.5
x1=3+2 square root of 3/5
x2=3- square root of 3/5
