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