We have the following quadratic function:
![f(x)=2x^2+7x-30](https://img.qammunity.org/2023/formulas/mathematics/college/bl06z8230hvhho24uroo4knknm0peb5ew2.png)
And we need to find its zeros i.e. the solutions to the equation:
![2x^2+7x-30=0](https://img.qammunity.org/2023/formulas/mathematics/college/as6os52vl4vt3n1sg95ciivswp46s08p7o.png)
Given a quadratic equation like the following:
![ax^2+bx+c=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/mvkhuzwnjhb4epaf7jjcoq2vi4zdi4350m.png)
Its solutions are given by the quadratic solving formula:
![x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/rxvf73usjbbwyik14knxdemoz21vfz2ufc.png)
In our case we have a=2, b=7 and c=-30. Then we get:
![x=\frac{-7\pm\sqrt[]{7^2-4\cdot2\cdot(-30)}}{2\cdot2}=\frac{-7\pm\sqrt[]{49+240}}{4}=\frac{-7\pm\sqrt[]{289}}{4}](https://img.qammunity.org/2023/formulas/mathematics/college/sbt7y7ntknyqw4e8hme3dmexf9smogntoh.png)
So we continue:
![x=\frac{-7\pm\sqrt[]{289}}{4}=(-7\pm17)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/jaoa9388xowua8bphlcr3o28bxs79lx8mw.png)
So we have two solutions:
![\begin{gathered} x_1=(-7+17)/(4)=(10)/(4)=2.5 \\ x_2=(-7-17)/(4)=-(24)/(4)=-6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o5wpik5emv218fm1dzq9ztoub5o2cd70st.png)
Then the answers are -6 and 2.5.