Solve for x
x\geq 10
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100-x^{2}+x^{2}\leq x+90
Consider \left(10-x\right)\left(x+10\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 10.
100\leq x+90
Combine -x^{2} and x^{2} to get 0.
x+90\geq 100
Swap sides so that all variable terms are on the left hand side. This changes the sign direction.
x\geq 100-90
Subtract 90 from both sides.
x\geq 10
Subtract 90 from 100 to get 10.
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