Question 623547: 3x^2=13x+56
3x^2-13x=56
then
x(3x-13)=56
x=56
3(56) -13. The result that i have is wrong; i do believe that i'm following the right process, but made a slight error, or maybe i missed a step. Your help is appreciated.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you're dealing with a quadratic equation so follow the rule for solving quadratic equations.
your equation is:
3x^2 = 13x + 56
subtract 13x and 56 from both sides of the equation to get:
3x^2 - 13x - 56 = 0
now your equation is in standard quadratic form.
this factors to be:
(3x+8) * (x-7) = 0
you get:
3x+8 = 0 and x-7 = 0 which results in:
x = -(8/3) or x = 7
plug those values in the original equation and the original equation should be true.
example:
when you replace x with -(8/3) your original equation becomes:
3*(-8/3)^2 = 13(-8/3) + 56
simplify this to get:
21.33333 = 21.33333
when you replace x with 7, your original equation becomes:
3*(7)^2 = 13*7 + 56
simplify this to get:
147 = 147
original equation is true (left side equals right side) in both cases so both answers satisfy the original equation.
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