Question

(3x-4)+20=5(x+1)^2+(2x-3)^2

Answer 1

(3x-4)+20=5(x+1)²+(2x+3)²
(3x-4)+20=5(x²+2×x×1+1²)+(2x²+2×2x×3+3²)
(3x-4)+20=5(x²+2x+1)+(4x²+12x+9)
3x-4+20=5x²+10x+5+4x²+12x+9
3x-4+20-5x²-10x-5-4x²-12x-9=0
-9x²-19x+2=0
a=-9
b=-19
c=2
∆=b²-4ac=(-19)²-4×(-9)×2=361+72=433
x(1,2)=(-(-19)±√433)/2×(-9)
x(1,2)=(19±√433)/-18
x(1)=(19+√433)/-18
x(2)=(19-√433)/-18