Question

3x^2+6x-7 what are the final two solutions

Answer 1

SOLVING:


`3X ^(2)+6X-7\quad\to\ Equalize\ the\ equation\ to\ zero.\\ \\3X ^(2)+6X-7=0`


The quadratic formula is:
`\boxed{X=\frac{-b\pm \sqrt{(b) ^(2)-4(a)(c) } }{2(a)}}`

Values:

a = 3
b = 6
c = - 7

STEPS:


`X= \frac{-(6)\pm \sqrt{(6) ^(2)-4(3)(-7) } }{2(3)}\\ \\ \\X= (-6\pm √(36+84) )/(6)\\ \\ \\X= (-6\pm √(120) )/(6)\\ \\ \\X= (-6\pm √(4*30) )/(6)\\ \\ \\X= \frac{-6\pm \sqrt{2 ^(2)*30 } }{6}\\ \\ \\X= (-6\pm2 √(30) )/(6)`


Getting X₁:


`X_(1)= (-6+2 √(30) )/(6)\to\ Factoring\ the\ numerator\ out\ common\ term\ 2.\\ \\ \\X_(1)= (2(-3+1 √(30)) )/(6)\to\ Reduce\ the\ common\ factor\ 2.\\ \\ \\X_(1)= (-3+1 √(30) )/(3)\\ \\ \\X_(1)=\boxed{ \boxed{ (-3+ √(30) )/(3) }}\quad\checkmark\ The\ first\ solution.`


Getting X₂:


`X_(2)= (-6-2 √(30) )/(6)\to\ Factoring\ the\ numerator\ out\ common\ term\ 2.\\ \\ \\X_(2)= (2(-3-1 √(30)) )/(6)\to\ Reduce\ the\ common\ factor\ 2.\\ \\ \\X_(2)= (-3-1 √(30) )/(3)\\ \\ \\X_(2)=\boxed{ \boxed{ (-3- √(30) )/(3)}}\quad\checkmark\ The\ second\ solution.`

GOOD LUCK...!!