## Algebra 1

x=-3, $\frac{5}{4}$
The quadratic formula states that if a$x^{2}$+bx+c=0, and a$\ne$0, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. If 4$x^{2}$+7x-15=0, then a=4, b=7, and c=-15, and to solve for x, then all we must do is plug the numbers into the quadratic formula then simplify. $x=\frac{-7\pm\sqrt{(7)^2-4(4)(-15)}}{2(4)}$ =$\frac{-7\pm\sqrt{49-(-240)}}{8}$ (7 squared=49, 4*4*-15=-240, and 2*4=8) =$\frac{-7\pm\sqrt{289}}{8}$ (49-(-240)=289) =$\frac{-7\pm17}{8}$ (The square root of 289=17) Since there is a plus minus symbol, we of course need to make sure that we compute both of our answers. x=$\frac{-7+17}{8}$=$\frac{10}{8}$=$\frac{5}{4}$ x=$\frac{-7-17}{8}$=$\frac{-24}{8}$=-3