Question

Can i please get some (help on my maths)

Answer 1

$ax^2+bx+c=0\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$


$We\ have:\\\\15x^2+7x-12=0\\\\a=15;\ b=7;\ c=-12\\\\substitute\\\\x_1=\dfrac{-7-\sqrt{7^2-4\cdot15\cdot(-12)}}{2\cdot15}=\dfrac{-7-\sqrt{49-5(15)(-12)}}{30}\\\\x_2=\dfrac{-7+\sqrt{49-4(15)(-12)}}{30}$

First option.