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\sqrt{7a-54}=-6+a
Subtract -a from both sides of the equation.
\left(\sqrt{7a-54}\right)^{2}=\left(-6+a\right)^{2}
Square both sides of the equation.
7a-54=\left(-6+a\right)^{2}
Calculate \sqrt{7a-54} to the power of 2 and get 7a-54.
7a-54=36-12a+a^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-6+a\right)^{2}.
7a-54-36=-12a+a^{2}
Subtract 36 from both sides.
7a-90=-12a+a^{2}
Subtract 36 from -54 to get -90.
7a-90+12a=a^{2}
Add 12a to both sides.
19a-90=a^{2}
Combine 7a and 12a to get 19a.
19a-90-a^{2}=0
Subtract a^{2} from both sides.
-a^{2}+19a-90=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=19 ab=-\left(-90\right)=90
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-90. To find a and b, set up a system to be solved.
1,90 2,45 3,30 5,18 6,15 9,10
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 90.
1+90=91 2+45=47 3+30=33 5+18=23 6+15=21 9+10=19
Calculate the sum for each pair.
a=10 b=9
The solution is the pair that gives sum 19.
\left(-a^{2}+10a\right)+\left(9a-90\right)
Rewrite -a^{2}+19a-90 as \left(-a^{2}+10a\right)+\left(9a-90\right).
-a\left(a-10\right)+9\left(a-10\right)
Factor out -a in the first and 9 in the second group.
\left(a-10\right)\left(-a+9\right)
Factor out common term a-10 by using distributive property.
a=10 a=9
To find equation solutions, solve a-10=0 and -a+9=0.
\sqrt{7\times 10-54}-10=-6
Substitute 10 for a in the equation \sqrt{7a-54}-a=-6.
-6=-6
Simplify. The value a=10 satisfies the equation.
\sqrt{7\times 9-54}-9=-6
Substitute 9 for a in the equation \sqrt{7a-54}-a=-6.
-6=-6
Simplify. The value a=9 satisfies the equation.
a=10 a=9
List all solutions of \sqrt{7a-54}=a-6.