Verified Answer :

AyaZ

\(2 x^{2} +x-10=0 \\ \\ (2x+5)(x-2)=0 \\ \\ x_{1} =- \frac{5}{2} \\ x_{2} =2\)

h(x)=2x2+x-10

2x²+x-10=0

(2x+5)(x-2)

x=-5/2
x=2


1)

h(x)=2x²+x-10
h(-5/2)= 2(-25/4)+[-5/2]-10
h(-5/2)=-25




2).

h(x)=2x²+x-10
h(2)=2(2)²+2-10
h(2)=8+(-8)
h(2)=0

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