Question 918851
first rewrite that b=2h+4 the write the formula for Area which is {{{A=(b*h)/2}}} then substitute and you get {{{A=h(2h+4)/2}}} then substitute {{{45=h(2h+4)/2}}} then {{{90=2h^2+4h}}} then {{{0=2h^2+4h-90}}} then *[invoke quadratic "x", 2, 4, -90 ] and then we only use the positive solution