Solve for x and find an approximate value

Question

$15 = \dfrac{((x+3)+(2x-3))h}{2}$
h = $(2x-3) -(x+3) $

& it is also given that $\sqrt{19}$ = 4.36

How can I simplify this & can you help me by explaining the steps

Answer 1

$$30 = ((2x-3)+(x+3))((2x-3)-(x+3))=(2x-3)^2-(x+3)^2=3x^2-18x$$ $$x^2-6x=10$$ $$x^2-6x+9=19$$ $$x=3\pm\sqrt {19}$$

Answer 2

$$15 = \dfrac{\left((x+3)+(2x-3)\right)h}{2}=\frac12 ((x+3)+(2x-3))\times((2x-3) -(x+3))=\frac12((2x-3)^2-(x+3)^2)=\frac12(3 x^2-18 x)$$ So, you need to solve $$30=3x^2-18x$$ I am sure that you can take it from here.