I will solve it by Lagrange multipliers, although there may be other possible ways, depending on what you've done in school.
Let
f(x,y)=25x^2+3y^2+20xy+10x+4y+10 (function to be minimized)
c(x,y)=5x^2+4xy-1 (constraint = 0)
we will set
p(x,y)=f(x,y)+λ*c(x,y)
where λ is the Lagrange multiplier (lambda), to be determined.
We will calculate the partial derivatives px, py and equate to zero.
Together with the constraint c(x,y)=0, we have three equations in three unknowns (x,y,λ).
Solving the 3 non-linear equations will give x=1/sqrt(5), y=0, and λ=-sqrt(5)-5.
The minimum value of f(x,y) is 15+2√5.
Here are the details:
p(x,y)=(4xy+5x^2-1)λ+3y^2+20xy+4y+25x^2+10x+10
px(x,y)=(4y+10x)λ+20y+50x+10 = 0
py(x,y)=4xλ+6y+20x+4 = 0
c(x,y)=4xy+5x^2-1 = 0
Solve for x, y & λ from the three non-linear equations above to get
x=1/sqrt(5), y=0, λ=-sqrt(5)-5.