SOLUTION: And the last one I'm stuck on, solve by completing the square. Thanks! :)
10x^2=4x+7
a. 2/5 +or- square root 74/5
b. 2/5 +or- square root 86/5
c. 1/5 +or- square root 74/10
Algebra ->
Square-cubic-other-roots
-> SOLUTION: And the last one I'm stuck on, solve by completing the square. Thanks! :)
10x^2=4x+7
a. 2/5 +or- square root 74/5
b. 2/5 +or- square root 86/5
c. 1/5 +or- square root 74/10
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Question 91729: And the last one I'm stuck on, solve by completing the square. Thanks! :)
10x^2=4x+7
a. 2/5 +or- square root 74/5
b. 2/5 +or- square root 86/5
c. 1/5 +or- square root 74/10 Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
10x^2=4x+7 ... 10x^2-4x-7=0
completing the square - step-by-step ... for ax^2+bx+c=0
1) by dividing or multiplying, set the coefficient of the squared term equal to one ... x^2+(b/a)x+(c/a)=0
2) by adding or subtracting, set the constant term (c/a) equal to zero ... x^2+(b/a)x=-(c/a)
3) add the square of one half of the x coefficient to both sides ... x^2+(b/a)x+(b/(2a))^2=-(c/a)+(b/(2a))^2
4) take the square root of both sides ... , realizing the root is positive AND negative
