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10a^2+16a-26=0
a = 10; b = 16; c = -26;
Δ = b2-4ac
Δ = 162-4·10·(-26)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-36}{2*10}=\frac{-52}{20} =-2+3/5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+36}{2*10}=\frac{20}{20} =1 $
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