Solve for x
x = \frac{7}{2} = 3\frac{1}{2} = 3.5
x = -\frac{7}{2} = -3\frac{1}{2} = -3.5
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\frac{176}{7}x^{2}=308
Multiply \frac{22}{7} and 8 to get \frac{176}{7}.
\frac{176}{7}x^{2}-308=0
Subtract 308 from both sides.
4x^{2}-49=0
Divide both sides by \frac{44}{7}.
\left(2x-7\right)\left(2x+7\right)=0
Consider 4x^{2}-49. Rewrite 4x^{2}-49 as \left(2x\right)^{2}-7^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).
x=\frac{7}{2} x=-\frac{7}{2}
To find equation solutions, solve 2x-7=0 and 2x+7=0.
\frac{176}{7}x^{2}=308
Multiply \frac{22}{7} and 8 to get \frac{176}{7}.
x^{2}=308\times \frac{7}{176}
Multiply both sides by \frac{7}{176}, the reciprocal of \frac{176}{7}.
x^{2}=\frac{49}{4}
Multiply 308 and \frac{7}{176} to get \frac{49}{4}.
x=\frac{7}{2} x=-\frac{7}{2}
Take the square root of both sides of the equation.
\frac{176}{7}x^{2}=308
Multiply \frac{22}{7} and 8 to get \frac{176}{7}.
\frac{176}{7}x^{2}-308=0
Subtract 308 from both sides.
x=\frac{0±\sqrt{0^{2}-4\times \frac{176}{7}\left(-308\right)}}{2\times \frac{176}{7}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{176}{7} for a, 0 for b, and -308 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\times \frac{176}{7}\left(-308\right)}}{2\times \frac{176}{7}}
Square 0.
x=\frac{0±\sqrt{-\frac{704}{7}\left(-308\right)}}{2\times \frac{176}{7}}
Multiply -4 times \frac{176}{7}.
x=\frac{0±\sqrt{30976}}{2\times \frac{176}{7}}
Multiply -\frac{704}{7} times -308.
x=\frac{0±176}{2\times \frac{176}{7}}
Take the square root of 30976.
x=\frac{0±176}{\frac{352}{7}}
Multiply 2 times \frac{176}{7}.
x=\frac{7}{2}
Now solve the equation x=\frac{0±176}{\frac{352}{7}} when ± is plus. Divide 176 by \frac{352}{7} by multiplying 176 by the reciprocal of \frac{352}{7}.
x=-\frac{7}{2}
Now solve the equation x=\frac{0±176}{\frac{352}{7}} when ± is minus. Divide -176 by \frac{352}{7} by multiplying -176 by the reciprocal of \frac{352}{7}.
x=\frac{7}{2} x=-\frac{7}{2}
The equation is now solved.
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Limits
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