Respuesta :
Add 10 to both sides so that the equation becomes 5x^2 + 11x - 2 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -11 ± √(11^2 - 4(5)(-2)) ] / ( 2(5) )
x = [ -11 ± √(121 - (-40) ) ] / ( 10 )
x = [ -11 ± √(161) ] / ( 10)
x = [ -11 ± sqrt(161) ] / ( 10 )
x = -11/10 ± sqrt(161)/10
The answers are -11/10 + sqrt(161)/10 and -11/10 - sqrt(161)/10.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -11 ± √(11^2 - 4(5)(-2)) ] / ( 2(5) )
x = [ -11 ± √(121 - (-40) ) ] / ( 10 )
x = [ -11 ± √(161) ] / ( 10)
x = [ -11 ± sqrt(161) ] / ( 10 )
x = -11/10 ± sqrt(161)/10
The answers are -11/10 + sqrt(161)/10 and -11/10 - sqrt(161)/10.
Add 10 and use the quadratic formula.
5x^2 +11x -2 = 0
x = (-11 ±√(11^2 -4(5)(-2)))/(2(5))
x = (-11 ±√161)/10 ≈ {-2.368858, 0.168858}
_____
The quadratic formula tells you that for [tex]ax^{2}+bx+c=0[/tex] the solutions are
[tex]x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex]
5x^2 +11x -2 = 0
x = (-11 ±√(11^2 -4(5)(-2)))/(2(5))
x = (-11 ±√161)/10 ≈ {-2.368858, 0.168858}
_____
The quadratic formula tells you that for [tex]ax^{2}+bx+c=0[/tex] the solutions are
[tex]x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex]
