We are given that f(a,b)=2a−3b2+7−5a2+b3 and f(k,−3)=−10. We want to find k.

To find k, we substitute a=k and b=−3 into the expression for f(a,b):

f(k,−3)=2(k)−3(−3)2+7−5(k)2+(−3)3

Simplifying the right side:

f(k,−3)=2k−27+7−5k2+27

Combining like terms:

f(k,−3)=−5k2+2k

We are given that f(k,−3)=−10, so we can set this equal to −10 and solve for k:

−5k2+2k=−10

Move all the terms to one side:

−5k2+2k+10=0

Factor the expression:

$ -(5k^2 - 2k - 10) = 0$

(k+2)(5k−5)=0

Therefore, either k+2=0 or 5k−5=0. Solving for k in each case gives us k=−2 or k=1.

However, we are given that f(k,−3)=−10. Let's evaluate f at each possible value of k:

If k=−2, then f(−2,−3)=2(−2)−3(−3)2+7−5(−2)2+(−3)3=16.

If k=1, then f(1,−3)=2(1)−3(−3)2+7−5(1)2+(−3)3=−10.

Since only f(1,−3)=−10, the solution for k is k=1.