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Solve by using quadratic formula: 5x^2+x=5
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We would like for this equation to be in the form \[ax^2 + bx + c = 0.\] That was we can see what a, b, and c are. First subtract 5 from both sides, \[5x^2+x-5=0\] Sometimes it helps to see the coefficient of 1, \[5x^2+1x-5=0\] Now a= 5, b = 1, c = -5. Plugging these values into the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] \[=\frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-5)}}{2(5)}\] \[=\frac{-1 \pm \sqrt{1 +20}}{10}\] \[=\frac{-1 \pm \sqrt{21}}{10}\] Now the square root of 21 does not simplify, so this is a final answer.
A= 5
not 1
oops
Thanks
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