Question

Can i please get some (help on my maths)

Answer 1

`ax^2+bx+c=0\\\\x_1=(-b-√(b^2-4ac))/(2a);\ x_2=(-b+√(b^2-4ac))/(2a)`



`We\ have:\\\\15x^2+7x-12=0\\\\a=15;\ b=7;\ c=-12\\\\substitute\\\\x_1=(-7-√(7^2-4\cdot15\cdot(-12)))/(2\cdot15)=(-7-√(49-5(15)(-12)))/(30)\\\\x_2=(-7+√(49-4(15)(-12)))/(30)`

First option.