Answer: \((2,0)\) and \(\left(10,-\frac{16}{5}\right)\).

From

\(2x+5y=4\)

we have

\(y=\frac{4-2x}{5}.\)

Substitute this into \(x^2+3xy=4\):

\(x^2+3x\left(\frac{4-2x}{5}\right)=4.\)

Multiplying by \(5\),

\(5x^2+12x-6x^2=20.\)

So

\(x^2-12x+20=0.\)

Factorising,

\((x-2)(x-10)=0.\)

Hence \(x=2\) or \(x=10\).

When \(x=2\),

\(2(2)+5y=4,\)

so \(y=0\).

When \(x=10\),

\(20+5y=4,\)

so \(5y=-16\), and hence \(y=-\frac{16}{5}\).

Therefore the solutions are

\((2,0)\quad\text{and}\quad\left(10,-\frac{16}{5}\right).\)